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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Similar questions

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

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

[N i (C N)_{4}]^{2 -}

[M n C l_{4}]^{2 -}

[Z n C l_{4}]^{2 -}

[N i (N H_{3})_{4}]^{2 +}

In the following molecules:
$C H_{3} B r, C H_{3} F, C H_{3} C l, C H_{3} I I, I I, I I I, I V$
The correct order of dipole moment will be:

\begin{matrix} [C o (H_{2} O)_{6}]^{3 +}; & [C o (C N)_{6}]^{3 -}; & [C o (I)_{6}]^{3 -}; & [C o (e n)_{3}]^{3 +} I & I I & I I I & I V \end{matrix}

F_{2} > C l_{2} > B r_{2} > I_{2}

F_{2} > C l_{2} > B r_{2} > I_{2}

C l > F > B r > I

F > C l > B r > I

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