Increasing paramganetism of $C a, A l, O, N$ is $C a < A l < O < N$ .
if true enter 1, else enter 0.
1

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Solution

The correct option is A 1
N has Z=7: $1 s^{2}, 2 s^{2} 2 p^{3}$ , so 5e in the outer shell , 1 pair(of 2e, from 2s2) and 3 unpaired electrons.
O has Z=8: $1 s^{2}, 2 s^{2} 2 p^{4}$ so 6e in the outer shell. 2 pairs and 2 unpaired ones (ready to form 2 bonds to achieve stability). There are 3 orbitals p, $p_{x}, p_{y}$ and $p_{z}$ . You have to first complete each one of them with 1e and if any electrons left, you pair them starting again from $p_{x}$ , until you fill the orbital up. So, in O case,4e go like this:2e in px, 1e in py and 1e in pz. This is why you have 2 unpaired e for O.
Al has Z=13 : $1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{1}$ so 3e in the outer shell, 3 unpaired electrons (1e in each of the $p_{x}$ , respectively $p_{y}$ and $p_{z}$ )

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{N a}^{+}, {C a}^{2 +}, {M g}^{2 +}, {A l}^{3 +}

{C a}^{2 +} < {N a}^{+} < {M g}^{2 +} < {A l}^{3 +}

A l < M g < N a

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{C a}^{2 +} < K^{+} < {C l}^{-} < S^{2 -}

C a O

C a : [A r] 4 s^{2} ⟶ C a^{2 +} [A r] O : 1 s^{2} 2 s^{2} 2 p^{4} ⟶ O^{2 -} 1 s^{2} 2 s^{2} 2 p^{6} C a^{2 +} [: ¨ O \cdot \cdot :]^{2 -}

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