<u>Answer:</u> The correct answer is Option D.
<u>Explanation:</u>
To calculate the hybridization of 
, we use the equation:
![\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5BV%2BN-C%2BA%5D)
where,
V = number of valence electrons present in central atom (S) = 6
N = number of monovalent atoms bonded to central atom = 0
C = charge of cation = 0
A = charge of anion = 0
Putting values in above equation, we get:
![\text{Number of electron pair}=\frac{1}{2}[6]=3](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20electron%20pair%7D%3D%5Cfrac%7B1%7D%7B2%7D%5B6%5D%3D3)
The number of electron pair around the central metal atom are 3. This means that the hybridization will be 
and the electronic geometry of the molecule will be trigonal planar.
Hence, the correct answer is Option D.
The correct answer is slow
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
