The correct statements among the following is/are

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$N O_{3}^{-}$ ion is stable but the corresponding $P O_{3}^{-}$ ion is less stable

$P O_{4}^{3 -}$ ion is stable but the corresponding $N O_{4}^{3 -}$ is ion is unstable

$N O_{3}^{-}$ ion is strong oxidizing agent but $P O_{4}^{3 -}$ cannot act as an oxidizing agent

$P O_{4}^{3 -}$ can act as reducing agent but $N O_{4}^{3 -}$ can act as strong oxidizing agent.

answer is A, B, C.

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Detailed Solution

Both $N O_{4}^{3 -}$ and $P O_{4}^{3 -}$ have 32 valence electrons, so both have similar Lewis structures. From the Lewis structures for $N O_{4}^{3 -}$ the central N atom has a tetrahedral arrangement of electron pairs. N is small atom. There is probability for steric hindrance by the 4 oxygen atoms around N atom. Further $N = O$ energy $(607 k J / m o l)$ is more than sum of two $N - O$ energies $2 \times 201 = 402 k J / m o l s$ . So $N O_{4}^{3 -}$ easily converts into stable $N O_{3}^{-}$

Phosphorous atom larger. No steric hindrance Further $P = O (544 k J / m o l)$ is smaller than the sum of two $P O (2 \times 335 = 670 k J / m o l)$ . So $P O_{4}^{3 -}$ is stable than $P O_{3}^{-}$

$P O_{3}^{-}$ and $N O_{3}^{-}$ each have 24 valence electrons, so both have similar Lewis structures. The planar structure which certain $P = O$ is not much stable as explained above.

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