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In $3 d$-transition series, which one has the least melting point?
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The correct answer is:
$\mathrm{Zn}$
The melting point of $d$-block elements is due to the presence of unpaired electrons in the last $d$-orbital or $s$-orbital. So, the element which has all the electrons in pairs will have the least melting point.
- Vanadium (V) with atomic number $23 .$ [Ar] $3 d^{3} 4 s^{2}$ has 3 unpaired electrons.
- Zinc (Zn) with atomic number (30). [Ar] $3 d^{10} 4 s^{2}$ has zero unpaired electron.
- Manganese (Mn) with atomic number $25 .$ $[\mathrm{Ar}] 3 d^{5} 4 s^{2}$ has five unpaired electrons.
- Copper (Cu) with atomic number 29 .
$[\mathrm{Ar}] 3 d^{10} 4 s^{1}$ has one unpaired electron.
Therefore zinc has least melting point.
- Vanadium (V) with atomic number $23 .$ [Ar] $3 d^{3} 4 s^{2}$ has 3 unpaired electrons.
- Zinc (Zn) with atomic number (30). [Ar] $3 d^{10} 4 s^{2}$ has zero unpaired electron.
- Manganese (Mn) with atomic number $25 .$ $[\mathrm{Ar}] 3 d^{5} 4 s^{2}$ has five unpaired electrons.
- Copper (Cu) with atomic number 29 .
$[\mathrm{Ar}] 3 d^{10} 4 s^{1}$ has one unpaired electron.
Therefore zinc has least melting point.
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