Explanation
1. $$p\pi-d\pi$$ bonding includes d-orbitals, if you observe the options only P belongs to 3rd period and has d-orbitals (though unoccupied in ground state)
2. Other options $$C$$ and $$N$$ belong to 2nd period and do not possess d orbitals.
A $$\sigma$$ bond is a type of covalent bond. It is formed by the head-on overlap of two atomic orbitals. These combinations of overlapping orbitals will be as $$ s-s, s-p_{z}\ or \ p_{z} - p_{z}$$.
Also, a $$\pi$$ bond is also a covalent bond. It is formed by the side-to-side overlap of two atomic orbitals. The atomic orbital may have combinations as $$ p_{x} - p_{x} or p_{y} - p_{y}$$
Thus as in the figure, total $$\sigma$$ bonds will be 17 and $$\pi$$ bonds will be 4.
Thus option B is correct.
Explanation:
Since the shape is square planar and all the bonds formed in this molecule are with the same element, thus the bonds are equal.
Since the shape is tetrahedral and all the bonds formed in this molecule are with the same element, thus the bonds are equal.
Since the type of bonds in this molecule are of two types i.e., $$C=C$$ and $$C-H$$. Hence in this molecule, all the bonds are not equal.
Final Answer:
Hence, the correct answer is option $$C.$$
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