The orbitals of same energy providing the most efficient overlapping are:

s p^{3} - s p^{3}

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s p - s p

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s p^{3} - s p^{2}

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all of the above

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Solution

The correct option is A $s p^{3} - s p^{3}$
The orbitals of same energy providing the most efficient overlapping are $s p^{3} - s p^{3}$ .
$s$ orbitals are closer to the nucleus than the $p$ orbitals. Hence, more is the $p$ character, more is the efficiency of overlap.

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Similar questions

N O_{3}^{-}, N O_{2}^{+} a n d N H_{4}^{+}

Most efficient overlapping is

(a) $s p^{2}$ − $s p^{2}$

(b) $s p^{3}$ − $s p^{3}$

(c)s−s

(d)sp−sp

Explain

C H_{4}, C_{2} H_{6}, a n d C_{2} H_{4}

N O_{3}^{-}, N O_{2}^{+}, N H_{4}^{+}

s p, s p^{2} a n d s p^{3}

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