Question

In the \(^{1}\)H NMR spectrum, multiplicity of the signal (bold and underlined H atom) in the following species is 
(I) \([\mathbf{H}\mathrm{Ni(OPEt_3)_4}]^{+}\)v 
(II) \(\mathrm{Ph_2Si(Me)\underline{\mathbf{H}}}\) 
(III) \(\mathrm{PH_3}\) 
(IV) \((\mathrm{Cp^*})_2\mathrm{Zr}\underline{\mathbf{H}}_2\) (Cp\(^*\)=pentamethylcyclopentadienyl)

• A. I- pentet, II- quartet, III- doublet and IV- singlet
• B. I- pentet, II- singlet, III- singlet and IV- doublet
• C. I- triplet, II- triplet, III- doublet and IV- doublet
• D. I- singlet, II- quartet, III- singlet and IV- singlet

Hint:

Equivalent nuclei never split each other.
Metal–hydrides often couple strongly to X (P, Si) nuclei; check the number of equivalent heteronuclei (\(n{+}1\) rule).
Low-abundance heteronuclei (e.g., \(^{29}\)Si) give satellite patterns; the main peak multiplicity is set by high-abundance neighbors.

Answer 1

The Correct Option is A

Step 1: Complex \([\mathrm{HNi(OPEt_3)_4}]^{+}\).  
The hydride couples to four equivalent \(^{31}\)P nuclei (\(I=\tfrac12\)). Thus \(n=4 \Rightarrow n{+}1=5\) lines: pentet. 
Step 2: \(\mathrm{Ph_2Si(Me)H}\). 
The Si–H proton is three bonds from the three equivalent \(\mathrm{Me}\) protons and shows \({}^{3}J_{H\text{–}H}\) coupling, giving a quartet. (Coupling to \(^{29}\)Si gives only weak satellite doublets because \(^{29}\)Si is 4.7% abundant.) 
Step 3: \(\mathrm{PH_3}\). 
Each H couples to one \(^{31}\)P nucleus (\(I=\tfrac12\)), giving a doublet. (H–H coupling is not observed among equivalent protons.) 
Step 4: \((\mathrm{Cp^*})_2\mathrm{ZrH_2\).} 
The two Zr–H hydrides are equivalent; \emph{equivalent} protons do not split each other, hence the hydride resonance is a singlet. 
\[ \boxed{\text{I pentet, II quartet, III doublet, IV singlet } \Rightarrow \text{ option (A).}} \]