What class of elements are providing the many electrons needed to make the H- ion?

What class of elements are providing the many electrons needed to make the H- ion?

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Continuous absorption by the H- ion reaches its maximum in the cool atmospheric layers of G stars. At these temperatures, most of the hydrogen atoms are neutral; but what class of elements is providing the many electrons needed to make the H- ion?

Easily ionised ones like sodium and potassium. Not much from lithium because that is rare.

You don't need many free electrons. The fraction of H$^{-}$/H is very small, something like $10^{-7}$ in the solar photosphere. And indeed the number densities of sodium and potassium to hydrogen are of that order.

Zinc group element

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Zinc group element, any of the four chemical elements that constitute Group 12 (IIb) of the periodic table—namely, zinc (Zn), cadmium (Cd), mercury (Hg), and copernicium (Cn). They have properties in common, but they also differ in significant respects. Zinc, cadmium, and mercury are metals with a silvery-white appearance and relatively low melting points and boiling points mercury is the only common metal that is liquid at room temperature, and its boiling point is lower than that of any other metal.

Three of these elements are found in different proportions in the Earth’s crust: it has been estimated that zinc is present to the extent of 80 parts per million (compared with 70 for copper and 16 for lead). The estimate for cadmium is only 0.15 commercially, it is always found associated with zinc or zinc–lead ores and is produced only as a by-product of zinc and lead smelting. The proportion of mercury in the Earth’s crust is estimated at 0.08 parts per million. All important mercury deposits consist of mercuric sulfide, known as the mineral cinnabar. Copernicium has only been produced in a particle accelerator.

The Same Everywhere

As far as we know, there are a limited number of basic elements. Up to this point in time, we have discovered or created about 120. Scientists just confirmed the creation of element 117 in 2014. While there are more elements to discover, the basic elements remain the same. Iron (Fe) atoms found on Earth are identical to iron atoms found on meteorites. The iron atoms in the red soil of Mars are also the same.

With the tools you learn here, you can explore and understand the Universe. You will never stop discovering new reactions and compounds, but the elements will be the same.

7.3 Lewis Symbols and Structures

Thus far in this chapter, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.

Lewis Symbols

We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:

Figure 7.9 shows the Lewis symbols for the elements of the third period of the periodic table.

Lewis symbols can also be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:

Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:

Figure 7.10 demonstrates the use of Lewis symbols to show the transfer of electrons during the formation of ionic compounds.

Lewis Structures

We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures , drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:

The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs ) and one shared pair of electrons (written between the atoms). A dash (or line) is sometimes used to indicate a shared pair of electrons:

A single shared pair of electrons is called a single bond . Each Cl atom interacts with eight valence electrons: the six in the lone pairs and the two in the single bond.

The Octet Rule

The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule .

The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons) this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule. The transition elements and inner transition elements also do not follow the octet rule:

Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:

Double and Triple Bonds

As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene):

A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN – ):

Writing Lewis Structures with the Octet Rule

For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:

For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:

  1. Determine the total number of valence (outer shell) electrons. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
  2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
  3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
  4. Place all remaining electrons on the central atom.
  5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.

    Determine the total number of valence (outer shell) electrons in the molecule or ion.
      For a molecule, we add the number of valence electrons on each atom in the molecule:

Example 7.4

Writing Lewis Structures


Check Your Learning


How Sciences Interconnect

Fullerene Chemistry

Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley (Figure 7.11), Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule (Figure 7.1). An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, shows promise in a variety of applications. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.

Exceptions to the Octet Rule

Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:

  • Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
  • Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
  • Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.

Odd-electron Molecules

We call molecules that contain an odd number of electrons free radicals . Nitric oxide, NO, is an example of an odd-electron molecule it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.

To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:

  1. Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
  2. Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N–O single bond:
  3. Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations thus oxygen has the filled valence shell:
  4. Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
  5. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)

Electron-deficient Molecules

We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13, outer atoms that are hydrogen, or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH2, and boron trifluoride, BF3, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF3, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule.

An atom like the boron atom in BF3, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom:

Hypervalent Molecules

Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules . Figure 7.12 shows the Lewis structures for two hypervalent molecules, PCl5 and SF6.

In some hypervalent molecules, such as IF5 and XeF4, some of the electrons in the outer shell of the central atom are lone pairs:

When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.


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Ionic Bonds

There are four types of bonds or interactions: ionic, covalent, hydrogen bonds, and van der Waals interactions. Ionic and covalent bonds are strong interactions that require a larger energy input to break apart. When an element donates an electron from its outer shell, as in the sodium atom example above, a positive ion is formed (Figure 2). The element accepting the electron is now negatively charged. Because positive and negative charges attract, these ions stay together and form an ionic bond, or a bond between ions. The elements bond together with the electron from one element staying predominantly with the other element. When Na + and Cl – ions combine to produce NaCl, an electron from a sodium atom stays with the other seven from the chlorine atom, and the sodium and chloride ions attract each other in a lattice of ions with a net zero charge.

Figure 2 In the formation of an ionic compound, metals lose electrons and nonmetals gain electrons to achieve an octet.

Find the Number of Electrons

For a neutral atom, the number of electrons is the same as the number of protons.

Often, the number of protons and electrons is not the same, so the atom carries a net positive or negative charge. You can determine the number of electrons in an ion if you know its charge. A cation carries a positive charge and has more protons than electrons. An anion carries a negative charge and has more electrons than protons. Neutrons do not have a net electric charge, so the number of neutrons does not matter in the calculation. The number of protons of an atom cannot change via any chemical reaction, so you add or subtract electrons to get the correct charge. If an ion has a 2+ charge, like Zn 2+ , this means there are two more protons than electrons.

If the ion has a 1- charge (simply written with a minus superscript), then there are more electrons than the number of protons. For F - , the number of protons (from the periodic table) is 9 and the number of electrons is:

Atom Diagrams Showing Electron Shell Configurations of the Elements

It's easier to understand electron configuration and valence if you can actually see the electrons surrounding atoms. For that, we have electron shell diagrams.

Here are electron shell atom diagrams for the elements, ordered by increasing atomic number.

For each electron shell atom diagram, the element symbol is listed in the nucleus. The electron shells are shown, moving outward from the nucleus. The final ring or shell of electrons contains the typical number of valence electrons for an atom of that element. The element atomic number and name are listed in the upper left. The upper right side shows the number of electrons in a neutral atom. Remember, a neutral atom contains the same number of protons and electrons.

The isotope is defined by the number of neutrons in an atom, which might be equal to the number of protons—or not.

An ion of an atom is one in which the number of protons and electrons is not the same. If there are more protons than electrons, an atomic ion has a positive charge and is called a cation. If there are more electrons than protons, the ion has a negative charge and is called an anion.

Elements are shown from atomic number 1 (hydrogen) up to 94 (plutonium). However, it's easy to determine the configuration of electrons for heavier elements by making a chart.

What class of elements are providing the many electrons needed to make the H- ion? - Astronomy

The elements can be classified as metals, nonmetals, or metalloids. Metals are good conductors of heat and electricity, and are malleable (they can be hammered into sheets) and ductile (they can be drawn into wire). Most of the metals are solids at room temperature, with a characteristic silvery shine (except for mercury, which is a liquid). Nonmetals are (usually) poor conductors of heat and electricity, and are not malleable or ductile many of the elemental nonmetals are gases at room temperature, while others are liquids and others are solids. The metalloids are intermediate in their properties. In their physical properties, they are more like the nonmetals, but under certain circumstances, several of them can be made to conduct electricity. These semiconductors are extremely important in computers and other electronic devices.

On many periodic tables, a jagged black line (see figure below) along the right side of the table separates the metals from the nonmetals. The metals are to the left of the line (except for hydrogen, which is a nonmetal), the nonmetals are to the right of the line, and the elements immediately adjacent to the line are the metalloids.

Structure of the Atom Class 9 Extra Questions Science Chapter 4

Extra Questions for Class 9 Science Chapter 4 Structure of the Atom

Structure of the Atom Class 9 Extra Questions Very Short Answer Questions

Question 1.
Which subatomic particle is absent in an ordinary hydrogen atom?

Question 2.
J. Chadwick discovered a subatomic particle which has no charge and has mass nearly equal to that of a proton. Name the particle and give its location in the atom.
The particle is neutron and it is present in the nucleus of the atom.

Question 3.
Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element. [NCERT Exemplar]
Yes, it is true for hydrogen atom which is represented as (_<1>^ <1>mathrm).

Question 4.
Electron attributes negative charge, protons attribute positive charge. An atom has both but why there is no charge?
The positive and negative charges of protons and electrons are equal in magnitude. So, atom as a whole is electrically neutral.

Question 5.
Write the electronic configuration of an element whose atomic number is 12.
K, L, M
2, 8, 2

Question 6.
What do you understand by ground state of an atom?
The state of an atom where all the electrons in the atom are in their lowest energy levels is called the ground state.

Question 7.
What is the maximum number of electrons which can be accommodated in ‘N’ shell?
N shell can accommodate maximum 32 electrons.

Question 8.
Write the correct representation of an element ‘X’ which contains 15 electrons and sixteen neutrons.
The correct representation of the element X is (_< 15 >^< 31 >< X >).

Question 9.
What will be the valency of an atom if it contains 3 protons and 4 neutrons?
The valency of the atom will be one.

Question 11.
Out of elements (_< 17 >^< 34 >< X >) and (_< 18 >^< 40 >< Y >), which is chemically more reactive and why?
The elements (_< 17 >^< 34 >< X >) is more reactive because its outermost shell is incomplete.

Question 12.
One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell? [NCERT Exemplar]
The charge would be +1.

Question 13.
In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed ? [NCERT Exemplar]
– 2.

Question 14.
Give two important applications of radioactive isotopes.

  • An isotope of carbon-12, C 14 , is used in carbon dating.
  • U 235 is used in the nuclear reactors to generate electricity.

Question 15.
Which isotope of hydrogen is present in heavy water?
Among the three isotopes of hydrogen, deuterium ((_<1>^ <2>mathrm))is found in heavy water.

Question 16.
Chemical formula of a metal sulphate is MSO4. What will be the formula of its chloride?

Question 17.
An element ‘A’ has valency +3, while another element ‘B’ has valency -2. Give the formula of their compound formed when ‘A’ reacts with ‘B’.
Element ‘A’ valency +3 (left)
Element ‘B’ valency – 2 (right)

Chemical formula = A2B3

Question 18.
Valency of an element X is 3. Write the chemical formula of its oxide.

Question 19.
Will 35 Cl and 37 Cl have different valencies? Justify your answer. [NCERT Exemplar]
No, 35 Cl and 37 Cl are isotopes of an element.

Question 20.
The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements? [NCERT Exemplar]

Structure of the Atom Class 9 Extra Questions Short Answer Questions—I

Question 1.
How do you know that nucleus is very small as compared to the size of atom?
Rutherford observed that when a-particles were bombarded on a very thin foil they bounced back. But the number of a-particles bouncing back got doubled when he doubled the thickness of gold foil. Then he concluded that the area of nucleus is very small in comparison to the total area of the atom.

Question 2.
Write two characteristics of the canal rays.

  • The canal rays are deflected by the magnetic fields in a direction opposite to that of the cathode rays.
  • They consist of positively charged particles.

Question 3.
Write the electronic configuration of a positively charged sodium ion (Na + ). Atomic number of sodium is 11.
Number of electrons in Na atom = Atomic number = 11
Number of electrons in Na + ion = 11 – 1 = 10
Electronic configuration of Na + ion: 2, 8

Question 4.
The electronic configuration of phosphorus atom is 2, 8, 5. Give the electronic configuration of P 3- ion.
Electronic configuration of P = 2, 8, 5
P atom gains 3e – to form P 3-
∴ P 3- has configuration = 2, 8, (5 + 3) = 2, 8, 8

Question 5.
The atomic number of Al and Cl are 13 and 17, respectively. What will be the number of electrons in Al 3+ and Cl – ?
Atomic number of Al = Number of electrons = 13
Number of electrons in Al 3+ = 13 – 3 = 10
Atomic number of chlorine = Number of electrons = 17
Number of electrons in Cl – = 17 + 1 = 18

Question 6.
Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17). [NCERT Exemplar]
The electronic distribution of Cl is 2, 8, 7. The L shell has eight electrons.

Question 7.
Define valence electrons. Which electrons of an atom are involved in the chemical bond formation with other atoms?
The electrons present in the outermost shell of an atom or ion are known as valence electrons.
In a chemical bond formation, only valence electrons of an atom take part.

Question 8.
Why do helium, neon and argon have a zero valency? [NCERT Exemplar]
Helium has two electrons in its energy shell, while argon and neon have 8 electrons in their valence shells. As these have maximum number of electrons in their valence shells, they do not have any tendency to combine with other elements. Hence, they have a valency equal to zero.

Question 9.
Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain. [NCERT Exemplar]
Helium atom has 2 electrons in its valence shell and its duplet is complete. Hence, the valency is zero.

Question 10.
Find out the valency of the atoms represented by the Figs. (a) and (b) [NCERT Exemplar]

(a) 0
(b) 1

Question 11.
Identify the Na + ion from the following figures. What is the valency of sodium atom? Give reason.

Figure number (ii) is correct because sodium ion (Na+) is formed when one electron is lost.
(egin> & ^<+>+1 mathrm^<->> <(2,8,1)>& <(2,8)>end)
The valency of sodium atom is one because stable (octet) electronic configuration is obtained after loss of one electron.

Question 12.
Calculate the number of neutrons present in the nucleus of an element X which is represented as (_< 15 >^< 31 >< X >) [NCERT Exemplar]
Mass number = No. of protons + No. of neutrons = 31
∴ Number of neutrons = 31 – Number of protons
= 31 – 15 = 16

Question 13.
Why do isotopes show similar chemical properties?
Isotopes have same atomic numbers and thus same number of electrons. Therefore, they have the same electronic configuration which provides them similar chemical properties.

Question 14.
An element ‘X’ has a valency 3(+):
(a) Write the formula of its phosphide.
(b) Write the formula of its carbonate.
(a) XP
(b) X2 (CO3)3

Question 15.
An element ‘Z’ forms the following compound when it reacts with hydrogen, chlorine, oxygen and phosphorous.
ZH3, ZCl3, Z2O3 and ZP
(a) What is the valency of element ‘Z’?
(6) Element ‘Z’ is metal or non-metal?
(a) The valency of ‘Z’ is 3.
(b) Element ‘Z’ is a metal because it is electropositive and is reacting with non-metals.

Structure of the Atom Class 9 Extra Questions Short Answer Questions-II

Question 1.
List any three distinguishing features between the models of an atom proposed by J.J. Thomson and Ernest Rutherford.

J. J. Thomson Model of Atom Rutherford’s Model
1. Positive charge forms a kernel. 1. Nucleus (positive charge) is in the centre.
2. Electrons present throughout the atom. 2. Electrons revolve in orbits.
3. No space is empty. 3.Most of the space is empty.

Question 2.
In the gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom,

1.00% of the a-particles were found to deflect at angles > 50°. If one mole of a-particles were bombarded on the gold foil, compute the number of a-particles that would deflect at angles less than 50°. [NCERT Exemplar]
% of α-particles deflected more than 50° = 1% of a-particles.
% of α-particles deflected less than 50° = 100 – 1 = 99%
Number of particles that deflected at an angle less than 50°
= (frac<99> <100> imes 6.022 imes 10^<23>)
= (frac<596.178> <100> imes 10^<23>)
= 5.96 × 10 23

Question 3.
Predict the valency of the following elements
(i) A (Atomic number 5)
(ii) B (Atomic number 12)
(iii) C (Atomic number 14)
(iv) D (Atomic number 17)
(i) Valency of element ‘A’ = 8 – 5 = 3
(ii) Valency of element ‘B’ = 12 – 10 = 2
(iii) Valency of element ‘C’= 14 – 10 = 4
(iv) Valency of element ‘D’= 18 – 17 = 1

Question 4.
An element ‘X’ contains 6 electrons in ‘M’ shell as valence electrons:
(a) What is the atomic number of ‘X’?
(b) Identify whether ‘X’ is a metal or non-metal.
(a) If ‘X’ contains 6 electrons in ‘M’ shell as valence electrons, then the electronic configuration of‘X’ is K = 2, L = 8, M = 6
∴ Atomic number = 16
(b) ‘X’ is a non-metal.

Question 5.
The atomic number of lithium is 3. Its mass number is 7.
(a) How many protons and neutrons are present in a lithium atom?
(b) Draw the diagram of a lithium atom.
(a) Number of neutrons = Mass number – atomic number
Number of neutrons = 7 – 3 = 4
Number of protons = atomic number
∴ Number of protons = 3
(b) Structure of a lithium atom

Question 6.
Complete the table on the basis of information available in the symbols given below [NCERT Exemplar]


Question 7.
In the atom of an element ‘Z’, 5 electrons are present in the outermost shell. It requires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? Write the formula of the compound which will be formed when ‘Z’ reacts with Na atom.
Number of electrons in the outermost shell = 5
Number of electrons required to make noble gas configuration = 8 – 5 = 3
The charge on the ion so formed = Z + 3e –
= Z 3-
The valency of Z = 3
Chemical formula of the compound:

Question 8.
(_< 86 >^< 222 >< Rn >) is an isotope of noble gas, radon. How many protons, neutrons and electrons are there in one atom of this radon isotope?
Atomic number of radon = 86
The number of protons = 86
The number of electrons = Number of protons
= 86
Number of neutrons = Atomic mass – Atomic number
= 222 – 86 = 136

Question 9.
What information do you get from the figures about the atomic number, valency of atoms X, Y and Z? Give your answer in a tabular form.


Question 10.
Write the molecular formulae for the following compounds:
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate [NCERT Exemplar]
(a) CuBr2
(b) Al(NO3)3
(c) Ca3(PO4)2
(d) Fe2S3
(e) HgCl2
(f) Mg(CH3COO)2

Question 11.
Write the molecular formulae of all the compounds that can be formed by the combination of following ions [NCERT Exemplar]
Cu 2+ , Na + , Fe 3+ , Cl – , (mathrm_<4>^<2->, mathrm_<4>^<3->)
CuCl2 CuSO4 Cu3 (PO4)2
NaCl Na2SO4 Na3PO4
FeCl3 Fe2(SO4)3 FePO4

Question 12.
Write the formula of the compounds formed by the following ions.
(a) Mg 2+ and S 2-
(b) Cu 2+ and OH
Name the compounds formed in each case.
(a) Ions Mg 2+ S 2-
Valencies 2 2
Compound: Mg2S2 or MgS Magnesium sulphate
(b) Ions Cu 2+ OH –
Valencies 2 1
Compound: Cu(OH)2 Copper (II) hydroxide.

Structure of the Atom Class 9 Extra Questions Long Answer Questions

Question 1.
(i) State the method of determining the valency of an element if its atomic number is given.
(ii) Determine the valency of the following elements, the atomic numbers of which are given in parenthesis:
Chlorine (17), Sulphur (16), Aluminium (13)
(i) The number of electrons gained, lost or shared to make the octet of electrons (in the outermost shell), gives us directly the combining capacity of the element, that is, the valency.

Question 2.
What is the gold foil experiment? Name the scientist who performed this experiment. Write the conclusions and shortcomings of Rutherford’s model of atom.
In 1911, Rutherford performed the gold foil experiment. He bombarded a stream of a-particles on a gold foil, a thin sheet which was 0.00006 cm thick in an evacuated chamber. An a-particle is a positively charged helium ion (He 2+ ). A simplified picture of this experiment is shown in the figure.

In this famous experiment, the following observations were made.

  • Most of the a-particles passed straight through the foil without any deflection. This concluded that most of the space inside of an atom is empty.
  • A few α-particles were deflected through small angle and few through larger angles. This happened due to positive charge on a-particles and core (nucleus) of the atom. The heavy positively charged ‘core’ was named as nucleus.
  • The number of α-particles which bounced back was very small. This concluded that the volume of the nucleus is very small in comparison to the total volume of the atom.

On the basis of gold foil experiment, Rutherford concluded that an atom consists of nucleus which has positive charge and it is surrounded with electrons which are moving around the nucleus. The number of electrons and protons are equal and the entire mass of the atom is concentrated at its nucleus.
Drawbacks in the Rutherford’s model

  • According to classical electro-magnetic theory, a moving charged particle, such as an electron under the influence of attractive force loses energy continuously in the form of radiations. As a result of this, electron should lose energy and therefore, should move in even smaller orbits ultimately falling into the nucleus. But the collapse does not occur. There is no explanation for this behaviour.
  • Rutherford did not specify the number of orbits and the number of electrons in each orbit.

Question 3.
In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model? [NCERT Exemplar]
Rutherford proposed a model in which electrons revolve around the nucleus in well-defined orbits. There is a positively charged centre in an atom called the nucleus. He also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly all the mass of an atom is centred in the nucleus. Whereas, Thomson proposed the model of an atom to be similar to a Christmas pudding. The electrons are studded like currants in a positively charged sphere like Christmas pudding and the mass of the atom was supposed to be uniformly distributed.

Question 4.
What are the postulates of Bohr’s model of an atom? [NCERT Exemplar]
The postulates put forth by Neils Bohr’s about the model of an atom:

  • Only certain special orbits known as discrete orbits of electrons, are allowed inside the atoms.
  • While revolving in discrete orbits the electrons do not radiate energy. These orbits are called energy levels. Energy levels in an atom are shown by circles.
    These orbits are represented by the letters K, L, M, N, … or the numbers n = 1, 2, 3, 4, ………..

Question 5.
The ratio of the radii of hydrogen atom and its nucleus is

10 5 . Assuming the atom and the nucleus to be spherical,
(i) what will be the ratio of their sizes?
(ii) If atom is represented by planet Earth ‘Re’ = 6.4 × 10 6 m. Estimate the size of the nucleus. [NCERT Exemplar]
(i) Volume of the sphere = (frac<4><3>) πr 3
Let R be the radius of the atom and r be that of the nucleus.

Question 6.
Show diagrammatically the electron distribution in a sodium atom and a sodium ion and also give their atomic number. [NCERT Exemplar] Answer:

Since the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na + ) is formed by the removal of one electron from a sodium atom. So, a sodium ion has 11 – 1= 10 electrons in it. Thus, electron distribution of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. Since, sodium atom and sodium ion contain the same number of protons, therefore, the atomic number of both is 11.

Question 7.
The given figure depicts the atomic structure of an atom of an element ‘X’.
Write the following information about the element ‘X’.

(a) Atomic number of ‘X’
(b) Atomic mass of ‘X’
(c) Valence electrons
(d) Valency of ‘X’
(e) ‘X’ should be metal or non-metal.
(a) Atomic number = Number of protons = 8
(ib) Atomic mass = Number of protons + Number of neutrons
= 8 + 10 = 18 u
(c) Valence electrons = 6
(d) Valency of ‘X’ = 8 – 6 = 2
(e) ‘X’ should be non-metal because there are six valence electrons hence it will take two more electrons to complete its outermost shell.

Structure of the Atom Class 9 Extra Questions HOTS (Higher Order Thinking Skills)

Question 1.
One electron is present in the outermost shell of the atom of an element ‘Z’.
(a) What will be the nature of this element?
(b) What will be the value of charge of the ion formed, if this electron is removed from the outermost shell?
(a) Element ‘Z’ will be a metal because it has only one electron in the outermost shell, so it is electropositive.
(b) After loss of one electron, ‘Z’ will acquire one positive charge.
Z → Z + + 1 e –

Question 2.
Composition of the nuclei of two atomic species ‘X’ and ‘Y’ are given below:

Give the mass number of ‘X’ and ‘Y’. What is the relationship between the two species?
(i) Atomic mass of element ‘X’ = Number of protons + Number of neutrons
= 8 + 8 = 16 u
(ii) Atomic mass of element ‘Y’ = Number of neutrons + Number of protons
= 10 + 8 = 18 u
Relationship between X and Y: The atomic number of both the elements is same but their atomic masses are different. Hence,they are isotopes of each other.

Question 3.
An atom ‘M’ of an element reacts with oxygen to form M2O3. Calculate the valency of the element ‘M’.
Two atoms of element ‘M’ combine with 3 atoms of oxygen.
∴ Number of oxygen atoms combining with one atom of element ‘M’ = (frac<3><2>)
Therefore, the valency of element ‘M’ = (frac<3><2>) × 2 = 3

Question 4.
Complete the following gaps in the given table:

We know that the number of protons = Atomic number
Number of protons = Number of electrons
Mass number = Number of protons + number of neutrons
Using these relationships, we can fill up these gaps as follows:

Question 5.
Explain why chlorine, whether as the element or its compounds, always has relative atomic mass of about 35.5.
The relative atomic mass is the average mass of one of the atoms and has to take into account the relative abundances of the various isotopes.
Natural chlorine always contains about (frac<3> <4> imes_<17>^ <35>mathrm ext < and >frac<1> <4> imes_<17>^ <37>mathrm).
Therefore, relative atomic mass of chlorine = (frac<3> <4> imes 35+frac<1> <4> imes 37)
= 35.5 u

Question 6.
An element ‘X’ has mass number 4 and atomic number 2. Write the valency of this element. Will it react with other atoms of different elements? [NCERT Exemplar]
We know that only valence electrons take part in bond formation with different atoms. In the atom of ‘X’ element there are only two electrons since atomic number is 2. Thus, K shell is fully filled for this atom. Hence, its valency is zero. It will not react with other atoms of different elements.

Watch the video: Periodensystem der Elemente I Teil 1 I musstewissen Chemie (January 2023).