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Solved Examples on Periodic Classification

Question 1:

Mg has a  higher melting point of a metal than Na. Explain

Solution

Melting point of a metal depends on the metallic bond strength. Mg has stronger metallic bonds due to

  • Its smaller size

  • Greater number of valence electrons involved in metallic bonding.

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Question 2:

Among fluorine – fluorine bond and chlorine – chlorine bond, which is more stronger and why?

Solution

In Cl—Cl bond, a filled p-orbital of chlorine can overlap with a suitable vacant d-orbital of adjacent chlorine thereby introducing some double bond character. Thus the bond strength increases. This is not possible in fluorine as it has got no vacant d-orbital.

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Question 3:

2nd electron affinity for halogens is zero – explain

Solution

After addition of one electron in halogen atom formation of halide ion takes place which has fully filled configuration (ns2np6) and thus possesses no tendency to gain one more electron.

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Question 4:

NaOH behaves as a base while Zn(OH)2 is amphoteric why?

Solution

In NaOH the bond electronegativity difference between Na and oxygen is greater than between H and O and therefore it is the Na–O bond that breaks releasing OH–. But in case of Zn—O—H bond the difference of electronegativity of Zn—O and O—H are almost same. So there is equal probability that the bond breaks in both ways leading to an amphoteric behaviour

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Question 5:

LiCl has a lower melting point than NaCl. Explain

Solution

Since LiCl is smaller in size than Na, it has a higher I.E. Therefore, it has a greater tendency to form covalent compounds. LiCl, being a covalent compound, has a lower melting point than NaCl which is ionic in nature.

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Question 6:

Why does xenon react with fluorine whereas neon does not?

Solution

Xe has a  lower ionisation energy than Ne. The valence electrons in Xe
(n = 5) are much farther from the nucleus than those of Ne (n=2) and much less tightly held by the nucleus; they are more willing to be shared that those in neon. Also xenon has empty 5d orbital which can help to accommodate the bonding pairs of electrons, while neon has all the valence orbitals filled.

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Question 7:

The electron affinity of chlorine is +37eV. How much energy in kJ is released when 3.55 g of chlorine is converted completely into Cl– ion in the gaseous state. (IeV = 96.45 kJ / mole)

Solution

Cl(g) + e–  →  Cl– (g) + 37 eV

Energy released when 1 mole (35.5 g) of chlorine atoms change completely into Cl–1(g) = 37 x  96.49 kJ / mol = 357.01 kJ

Energy released when 3.55 g of chlorine atoms change completely into Cl– (g) = (357.01/35.5) x 3.55 = 35.701 kJ

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Question 8:

Compare qualitatively the first and second I.P. of Cu & Zn. Explain the observation.

Solution

First I.P. of Cu is less than that of Zn because removal of one electron from 4s orbital of  Cu leaves completely filled orbitals and in that of Zn, a completely filled 4s orbital is converted to half-filled orbital. But in case of 2nd I.P. the case is reversed as Cu+ gives an incompletely filled orbital (3d9) and Zn+ gives completely filled (3d10) orbital.

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Question 9:

The electron affinity of sulfur is greater than oxygen. Why?

Solution

This is because of smaller size of oxygen due to which it has got higher change density and thus electronic repulsion increases as it takes electron. So its E.A. is less than sulphur.

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Question 10:

In s & p block elements the oxidation state change by 2 units but in transition elements it changes in units of 1. Explain.

Solution

This is due to a pair of electrons remaining paired in –ous form and becoming unpaired in –ic form (e.g. Sn, P, Te etc.]. But in transition metals different no. of d-electrons may take part in bonding [e.g. Fe2+, Fe3+ & Cu+, Cu2+ etc.].

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Question 11:

Why is the atomic radius of oxygen slightly more than that of nitrogen?

Solution

Nitrogen atom has a half-filled sub-shell 2p3; the next electron in oxygen enters an occupied p-orbital (2p4); so the 2pz orbital has a pair of electrons. The resulting repulsion between this pair of electrons more than offsets the attraction of the increased nuclear charge. This can be seen from the values of the atomic radii given in the table.

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Question 12:

Which one in larger size and why?

  • Na or Na+;

  • S or S2–‑;

  • K+ or Cl–

Solution

  • When an electron is removed from an atom, the effective nuclear charge increases. Hence, the electron cloud shrinks. Na > Na+

  • When one or more electrons are added to a neutral atom, the nuclear charge has to attract the additional electrons also. Hence, effective nuclear charge decreases. S2– > S

  • K+ and Cl– are isoelectronic, that is, they possess 18 electrons. In the case of K+, there are 19P attracking the 18 electrons while in Cl– there are only 17P attracting the 18 electrons. Hence, effective nuclear charge is greater in K+.Cl–  > K+

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Question 13:

Be has higher ionisation potential than B – explain.

Solution

Be has fully filled orbital (1s22s2) but boron does not have a fully filled orbital (1s22s22p1).  Again s electron is more penetrating compared to p-electron. That’s why more energy is required to remove electron from 2s orbital compared to 2p orbital. So Be has higher ionisation energy.

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Question 14:

The EA of Mg is negative while that of Cl is positive. Account for the observation.

Solution

Mg has the electronic configuration 1s22s22p63s2. It has a tendency to lose electrons in order to attain a stable octet configuration. When an electron is introduced into its outermost shell, energy has to be supplied, i.e. it is an endothermic process. EA is assigned a negative value. Chlorine has the electronic configuration 1s22s22p63s23p5. It readily accepts an electron. Hence, the electronation process is exothermic and its EA is positive.

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