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In this question my answer was (A) because I thought $\ce{NH2}$ group would give more electrons than $\ce{Cl-}$. But the correct answer is (D) and my teacher's reason is that chlorine has more negative Inductive effect than positive Resonance effect. Is that true?
Chloro group is considered a deactivating group according to Hammett plots of each possible reaction. The $\sigma_\mathrm{para}$ of $\ce{Cl}$-substituent is listed as $+0.227$ while $\sigma_\mathrm{meta}$ of $\ce{Cl}$ is listed as $+0.373$. It is known fact that $\sigma_\mathrm{meta}$ is an indicative of how much inductive effect contribute to the reaction, while $\sigma_\mathrm{para}$ is a combination of the strength of both inductive and mesomeric effects (more resonance effect than inductive because the substitution is one carbon away compared to meta-position). For comparison, when the substitution is $\ce{H}$, both effects are zero while when the substitution is $\ce{NO2}$ group, $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ are $+0.710$ and $+0.778$, respectively (resonance effect predominates). Based on Hammett's original definition, when $\sigma$ of any group has positive sign,it is electron withdrawing group (EWG) and the numerical value is an indication of hoe big electron withdrawing ability compared to that of $\ce{H}$ (value of which is always zero). Note that Hammett has assigned these sign and value according to his research on the acidity of substituted aromatic carboxylic acids (EWG substitution makes acid stronger).
Contrary to carboxylic acids, EWG substitution at aromatic nucleus makes amino bases weaker because EWG dilutes the electron density on active $\ce{N}$ atom by inductive and/or resonance effects. Therefore, largere the positive number of $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of EWG, weaker the base. Also keep in mind that $\sigma_\mathrm{meta}$ and $\sigma_\mathrm{para}$ of $\ce{NH2}$ group are $-0.161$ and $-0.66$, respectively. Hence, basic strength of given benzamidine are decreased in order of their 4-substitution of $\ce{NH2} \gt \ce{H} \gt \ce{Cl} \gt \ce{NO2}$ meaning your correct answer is $\bf{(D)}$. It is noteworthy that this order remains the same even if these substitutions are at 3-position of the aromatic nucleus (based on $\sigma_\mathrm{meta}$ values).
I can't really find the $\mathrm{p}K_\mathrm{a}$ of relevant compounds. However, I found $\mathrm{p}K_\mathrm{a}$ values of some guanidines, $\ce{X-C6H4-N=N(N(CH3)2)N(N(CH3)2}$ (parent compound: 2-phenyl-1,1,3,3-tetramethylguanidine, $\ce{C6H5-N=N(N(CH3)2)N(N(CH3)2}$) which would prove my point (Ref.1):
$$ \begin{array}{c|ccc} \hline \ce{X} & \sigma_{(m,p)}^a & \mathrm{p}K_\mathrm{a} \ \text{of 2-Ar-guanidine} & \mathrm{p}K_\mathrm{a} \ \text{of aniline}^b \\ \hline \ce{H} & 0 & 12.18 & 4.62 \\ \text{3-}\ce{CH3} & -0.06 & 12.25 & 4.70 \\ \text{4-}\ce{CH3} & -0.14 & 12.37 & 5.11 \\ \text{3-}\ce{OCH3} & 0.11 & 11.96 & 4.23 \\ \text{4-}\ce{OCH3} & -0.28 & 12.57 & 5.34 \\ \text{3-}\ce{Cl} & 0.37 & 11.47 & 3.52 \\ \text{4-}\ce{Cl} & 0.24 & 11.70 & 3.98 \\ \text{4-}\ce{NO2} & 1.26 & 9.78 & 0.99 \\ \hline \end{array}\\ ^a: \text{Ref.2 and } ^b: \text{Ref.3.} $$
References:
Yes as you said the inductive effect of chlorine exceeds the strength of it's +R effect.
This effect becomes clear when you study aromaticity, that chlorine is a deactivating group (i.e it destabilizes the carbocation formed) but is still Ortho - para directing, because it still causes stabilization by the +R effect in those positions.
However unlike other +R groups which are also activating, chlorine is deactivating, because its inductive effect is slightly stronger than it's +R effect
( Note that all comparisons are based on which effect causes more stability)
