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Question

Which of the following statements are correct?
(I) Both melting and boiling points of H2O are higher than those of H2Te .
(II) In both N2O5 and N2O4 all N−O bond lengths are equivalent.
(III) In both crystalline NaHCO3 and KHCO3.HCO−3 forms only dimeric anion through hydrogen bond.
(IV) B2,C2,N2 on further ionization (removal of single electron) form thermodynamically less stable species.

A
(I) and (II)
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B
(III) and (IV)
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C
(II) and (III)
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D
(I) and (IV)
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Solution

The correct option is D (I) and (IV)
(I) On account of Hbonding H2O has higher melting and boiling points.
H2OH2Tem.p/K273222b.p/K373269

(II) (III) (a) In NaHCO3, the HCO3 ions are linked into an infinite chain and (b) in KHCO3,HCO3 forms a dimeric anion.

(IV) Each molecule through ionization (after removal of single electron) corresponding cation is formed.
B2B.O=1 and B+2B.O=12C2B.O.=2 and C+2B.O.=1.5
N2B.O=3 and N+2B.O=2.5
As , bond strength bond order.
So , with ionization bond strength decreases which eventually decreases thermal stability of the molecule.

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