Choose the correct order of bond strength by overlapping of atomic orbitals:
A
1s−1s>1s−2s>1s−2p
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B
2s−2s>2s−2p>2p−2p
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C
2s−2p>2s−2s>2p−2p
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D
1s−1s>1s−2p>1s−2s
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Solution
The correct option is D1s−1s>1s−2p>1s−2s When n is less, then overlapping is stronger. Therefore, the bond formed by 1s−1s overlap is strongest. When n is different value, overlap with directional nature domainants. Hence 1s−2p>1s−2s