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The products formed in the above reaction are
Reason: $He$ is soluble in blood.
| List $I$ Complex | List $II$ Crystal Field splitting energy $\left(\Delta_0\right)$ |
| $A$ $\left[ Ti \left( H _2 O \right)_6\right]^{2+}$ | $I$ $-1.2$ |
| $B$ $\left[ V \left( H _2 O \right)_6\right]^{2+}$ | $II$ $-0.6$ |
| $C$ $\left[ Mn \left( H _2 O \right)_6\right]^{3+}$ | $III$ $0$ |
| $D$ $\left[ Fe \left( H _2 O \right)_6\right]^{3+}$ | $IV$ $-0.8$ |
$(A)$ The acidity of compound $I$ is due to delocalization in the conjugate base.
$(B)$ The conjugate base of compound IV is aromatic.
$(C)$ Compound II becomes more acidic, when it has a $- NO _2$ substituent.
$(D)$ The acidity of compounds follows the order $I > IV > V > II > III$.