Correct Answer - `{:(3,5,2,4),(P,Q,R,S):}`
`NO^(-)[BF_4]^-`
(P)Molecular orbital electronic configuration ,
`NO^(+)` is derivative of `O_2`, so MOT configuration is :
`sigma1s^2 sigma^(**)1s^2sigma2s^2 sigma^(**)2s^2sigma2p_x^2 pi2p_y^2=pi2p_z^2` ( x is taken as molecular axis)
Bond order =`(10-4)/2=3`
One `sigma` bond are two `pi` bonds.
(Q)Number of `sigma` bonds in `NO^+` is one and in `BF_4^-` are four. So total number of `sigma` bonds are five.
(R)Number of `pi` bonds is `NO^+` are two.
(S)
Steric number of central atom boron is 4+0=4, so its hybridisation is `sp^3` and thus the number of hybrid orbitals involved in `sp^3` hybridisation is four.
