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Question Number 23146 by Tinkutara last updated on 26/Oct/17

$$\mathrm{Square}\mathrm{planar}\mathrm{complex}\mathrm{is}\mathrm{formed}\mathrm{by}$$$$\mathrm{hybridisation}\mathrm{of}\mathrm{which}\mathrm{atomic}\mathrm{orbitals}?$$$$\left(1\right)s,p_x,p_y,p_z$$$$\left(2\right)s,p_x,p_y,d_{z^2}$$$$\left(3\right)s,p_x,p_y,d_{x^2-y^2}$$$$\left(4\right)s,p_x,p_y,d_{z^3}$$

Commented by math solver last updated on 26/Oct/17

$$\left(3\right)?\mathrm{although}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{confused}\mathrm{in}$$$$2\mathrm{and}3\mathrm{option}$$$$$$

Commented by math solver last updated on 26/Oct/17

$$\mathrm{actually}\mathrm{square}\mathrm{planar}\mathrm{compex}$$$$\mathrm{has}\mathrm{hybridisation}{\mathrm{dsp}}^2.$$$$\mathrm{so}\mathrm{going}\mathrm{by}\mathrm{option}\mathrm{only}2\mathrm{and}3$$$$\mathrm{satisfy}\mathrm{and}{\mathrm{d}}_{{\mathrm{z}}_{}^3}\mathrm{is}\mathrm{nothing}....$$$$$$

Commented by Tinkutara last updated on 26/Oct/17

$${\mathrm{Doesn}}^{\prime }\mathrm{t}\mathrm{square}\mathrm{planar}\mathrm{has}{sp}^3?\mathrm{This}$$$$\mathrm{means}4\sigma \mathrm{bonds}\mathrm{and}0\mathrm{lone}\mathrm{pairs}.\mathrm{And}$$$$\mathrm{answer}\mathrm{is}\mathrm{only}\left(3\right).$$

Commented by math solver last updated on 26/Oct/17

$$\mathrm{square}\mathrm{planar}\mathrm{and}\mathrm{square}\mathrm{planar}$$$$\mathrm{complex}\mathrm{are}\mathrm{different}.$$$$\mathrm{square}\mathrm{planar}\mathrm{has}\mathrm{hybridization}$$$${\mathrm{sp}}^3{\mathrm{d}}^2.\mathrm{eg}:{\mathrm{XeF}}_4.$$

Commented by Tinkutara last updated on 27/Oct/17

$${\mathrm{XeF}}_4\mathrm{has}2\mathrm{axial}\mathrm{lone}\mathrm{pairs}\mathrm{also}.\mathrm{So}\mathrm{why}$$$$\mathrm{this}\mathrm{is}\mathrm{square}\mathrm{planar}?\mathrm{It}\mathrm{should}\mathrm{be}$$$$\mathrm{bipyramidal}.$$

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