- AIron (Z = 26)
- BChromium (Z = 24)
- CVanadium (Z = 23)
- DScandium (Z = 21)
Explanation:
Scandium(Z = 21) forms colourless compounds.
Scandium(Z = 21) has valence shell electron configuration of 3d14s2
In its +3 oxidation state, it looses 3 electrons and has valence shell electron configuration of 3d°4s°. Since no unpaired electron is present, it forms colourless compounds.
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$Q (47 \%)$ and $R (2 \%)$. The major product(s) of the following reaction sequence is (are)
$\begin{array}{l}\begin{array}{ll}\text { 1) } Ac _2 O \text {, pyridine } \quad \text { 1) } Sn / HCl \end{array} \\ R \xrightarrow{2) Br _2, CH _3 CO _2 H } S \quad \xrightarrow{2) Br _2 / H _2 O \text { (excess) }} \text { major product(s) } \\ \text { 3) } H _3 O ^{+} \quad \text {3) } NaNO _2, HCl / 273-278 K \\ \text { 4) } NaNO _2, HCl / 273-278 K \quad \text { 4) } H _3 PO _2 \\ \text { 5) } EtOH , \Delta \\\end{array}$
$\left( C _6 H _5\right)_3 C - Cl \frac{ OH ^{-}}{\text {Pyridine }}\left( C _6 H _5\right)_3 C - OH$
$CuI_{(s)} + e^{\circleddash} \to Cu_{(s)} + I^{\circleddash}_{(aq)}$ ; $E^o = -0.17\, V$
$Zn^{+2}_{(aq)} + 2e^{\circleddash} \to Zn_{(s)}$ ; $E^o = -0.76\, V$
