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Assertion : The 1^(st) IP of Be is great...

Assertion : The `1^(st)` IP of Be is greater than that of B
Reason : 2p orbital is lower in energy than 2s

A

If both (A) and (R) are correct, but (R) is not the correct explanation of (A).

B

If both (A) and (R) are correct, but (R) is not the correct explanation of (A)

C

If (A) is correct, but (R) is incorrect.

D

If both (A) and (R) are incorrect.

Text Solution

Verified by Experts

The correct Answer is:
C
`IE_(1)` of Be is greater than `IE_(1)` of B because of fully filled configuration. 2p orbital has higher energy as it is at more distance from nucleus in comparison to 2s orbital.
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