MCQ
As a result of $sp$ hybridization, we get
- ATwo mutual perpendicular orbitals
- ✓Two orbitals at ${180^o}$
- CFour orbitals in tetrahedral directions
- DThree orbitals in the same plane
✓
Answer
Correct option: B.
Two orbitals at ${180^o}$
(b) $sp-$hybridization gives two orbitals at ${180^o}$ with Linear structure.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the concentration of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ in blood is $0.72\, \mathrm{~g} \,\mathrm{~L}^{-1}$, the molarity of glucose in blood is $.....\,\times 10^{-3} \mathrm{M}$. (Nearest integer)
→(Given: Atomic mass of $\mathrm{C}=12, \mathrm{H}=1, \mathrm{O}=16 \mathrm{u}$ )
An organic dye, cosine used to detect end point of precipitation titration by adsorption is called
What is the molar solubility of $Fe(OH)_2$ $(K_{sp} = 8.0 \times 10^{-16})$ at $pH\,\,13.0$ ?
→$\dot{ Cl }+ CH _{4} \rightarrow A + B$
→$A$ and $B$ in the above atmospheric reaction step are.
To Convert cyclononatetraene in aromatic ion the process needed is
$0.01\, M\, KMnO _{4}$ solution was added to $20.0\, mL$ of $0.05\, M$ Mohr's salt solution through a burette. The initial reading of $50\, mL$ burette is zero. The volume of $KMnO _{4}$ solution left in the burette after the end point is $.....\,mL$. (nearest integer)
→The product $(A)$ of given alkoxymercuration de-mercuration is
→The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds:
The hottest region of Bunsen flame shown in the figure below is :
For the change ${H_2}O(l) \to {H_2}O(g);\,P = 1\,atm$, $T = 373\,K$, the free energy change $\Delta G = 0$. This indicates that
→