Download App

STD 11 - 4. Chemical bonding and molecular structure — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 4. Chemical bonding and molecular structure4 Marks
MCQ
A square planar complex is formed by hybridisation of which atomic orbitals
  • A
    $s,\,\,{p_x},\,\,{p_y},\,\,{d_{yz}}$
  • ✓
    $s,\,\,{p_x},\,\,{p_y},\,\,{d_{{x^2} - {y^2}}}$
  • C
    $s,\,\,{p_x},\,\,{p_{y,\,\,}}{d_{{z^2}}}$
  • D
    $s,\,\,{p_y},\,\,{p_z},{d_{xy}}$
✓

Answer

Correct option: B.
$s,\,\,{p_x},\,\,{p_y},\,\,{d_{{x^2} - {y^2}}}$
(b)For square planar geometry hybridization is $ds{p^2}$involving $s$,${p_x},\,{p_y}$ and ${d_{{x^2} - }}_{{y^2}}$ orbital.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

NEET STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The compound which on reaction with aqueous nitrous acid on $HN{O_2}$ at low temperature produces an oily nitrosoamine is
→
Consider the following reaction

${R_2}SiC{l_2} + {H_2}O \to (1)\xrightarrow{{polymerisation}}(2)$ 

Compound $(2)$ in above reaction is

→
A reaction

$2NO + 2H_2 \longrightarrow N_2 + 2H_2O$

has following mechanism

Step $-I$ : $2NO \longrightarrow N_2O_2$

Step $-II$ : $N_2O_2 + H_2 \longrightarrow N_2O + H_2O$

Step $-III$ : $N_2O + H_2 \longrightarrow N_2 + H_2O$

Which of the following substance is a reaction intermediate

→
Which molecule does not show zero dipole moment
→
If $K_{sp}$ of $CaF_2$ at $25\,^oC$ is $1 .7 \times 10^{-10},$ the combination amongst the following which gives a precipitate of $CaF_2$ is
→
The correct statements among $I$ to $III$ are

$I$. Valence bond theory cannot explain the color exhibited by transition metal complexes

$II$. Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes

$III$. Valence bond theory cannot distinguish ligands as weak and strong field ones

→
The shape of $XeO _2 F _2$ molecule is
→
The potassium ferrocyanide solution gives Prussian blue colour, when added to :
→
When isobutane is treated with $Cl_2$ in sunlight, then major product is
→
Compound $A\,(C_9H_{10}O)$ shows positive iodoform test. Oxidation of $A$ with $KMnO_4/KOH$ gives acid $B(C_8H_6O_4)$. Anhydride of $B$ is used for the preparation of phenolphthalein. Compound $A$ is
→