Which of the following is correct with respect to energy for multi-electron system?

$I . 1 s < 2 s I I . 2 p < 3 p I I I . 3 d_{x y} = 3 d_{y z} I V . 3 d > 4 s$

I, II

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I, II, III

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I, II, IV

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I, II, III and IV

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Solution

The correct option is D
I, II, III and IV

Let's discuss one by one.

$1 s < 2 s \Rightarrow$ Because same 'l' but 'n' value is more for 2s.

$2 p < 3 p \Rightarrow$ same reason

$3 d_{x y} = 3 d_{y z} \Rightarrow$ Because all d-orbitals will be denigrate in a particular energy level.

$3 d > 4 s \Rightarrow (n + l) f o r 3 d = 3 + 2 = 5$
$(n + l) f o r 4 s = 4 + 0 = 4$

This is because of penetration power.

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Which of the following is correct with respect to energy for multi-electron system? I. 1s<2s II. 2p<3pIII. 3dxy=3dyzIV. 3d>4s

Which of the following is correct with respect to energy for multi-electron system?

$I . 1 s < 2 s I I . 2 p < 3 p I I I . 3 d_{x y} = 3 d_{y z} I V . 3 d > 4 s$