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
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The end product $(C)$ is
Product of the above reaction will be
$Zn(s) + Cu^{2+}(M) \to Zn^{2+}(M') + Cu(s);$ $E^o _{cell} = 1.10\,V$
$X$-axis :$log_{10}$ $\frac{{[Z{n^{2 + }}]}}{{[C{u^{2 + }}]}}$, $Y$ -axis : $E_{cell}$