Identify the correct statement(s) for $\mathrm{H}_{3}\mathrm{B}.\mathrm{CO}$(A) $\mathrm{sp}^{2}$ hybridized orbital of B accepts the lone pair of CO.(B) Its $\bar{v}_{\mathrm{CO}}$ value is more than that for free CO(C) Formal oxidation state of C is +4 in the compoundAnswer is
(a)A and B
(b)B only
(c)A only
(d)A and C
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
In H₃B·CO the boron is sp³ (not sp²), ν(CO) (2165 cm⁻¹) is higher than free CO because there is no back-donation, and carbon stays formally +2. Only B is correct.