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Question: In analogy to\(O_{2}^{+}\lbrack PtF_{6}\rbrack^{-}\), a compound\(N_{2}^{+}\lbrack PtF_{6}\rbrack^{-...

In analogy toO2+[PtF6]−O_{2}^{+}\lbrack PtF_{6}\rbrack^{-}, a compoundN2+[PtF6]−N_{2}^{+}\lbrack PtF_{6}\rbrack^{-} will not be formed because –

A

The IE ofN2N_{2} gas is higher than that of N-atom

B

The IE ofN2N_{2}gas is lower than that ofO2O_{2}gas

C

The IE ofN2N_{2}gas is higher than that ofO2O_{2}gas

D

None of these

Answer

The IE ofN2N_{2}gas is higher than that ofO2O_{2}gas

Explanation

Solution

The IE ofN2N_{2}gas (1503 kJmol−1mol^{- 1})is higher than that of O2O_{2} gas (1175 kJ mol−1mol^{- 1} and it cannot lose its electron easily)