For the folllowing molecules the correct order of bond angles is:

x > y > z

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y > z > x

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z > x > y

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z > y > x

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Solution

The correct option is A x > y > z
$∙$ VSEPR theory determines the shape of a molecule based on the repulsions between bond pair-bond pair, bond pair-lone pair and lone pair-lone pair electrons on the central atom.

$∙$ In a molecule repulsion between electron pairs follow the order:
lone pair - lone pair > lone pair - bond pair > bond pair - bond pair

$∙$ If a lone pair is present on the central atom, the bond angle decreases because the distance between the lone pair and bond pair electrons increases to minimize repulsions.

$∙$ If multiple lone pairs are present on the central atom the angle decreases even more due to stronger repulsions.

$∙$ Therefore bond angle z is least due to the presence of two lone pair followed by y (one lone pair) and x (no lone pairs).

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\frac{x (y + z - x)}{log x} = \frac{y (z + x - y)}{log y} = \frac{z (z + x - y)}{log z},

x^{y} y^{x} = z^{y} y^{z} = x^{z} z^{x}

∣ ∣ ∣ ∣ \begin{matrix} 3 x & - x + y & - x + z x - y & 3 y & z - y x - z & y - z & 3 z \end{matrix} ∣ ∣ ∣ ∣ = 3 (x + y + z) (x y + y z + x z)

x^{2} + x y + x z = 18

y^{2} + y z + y z + 12 = 0

z^{2} + z x + z y = 30

x^{2} + x y + x z = 135, y^{2} + y z + x y = 351

z^{2} + z x + z y = 243

(x, y, z > 0)

x^{2} + y^{2} + z^{2}

x^{2} + x y + x z = 135, y^{2} + y z + y x = 351

z^{2} + z x + z y = 243

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