Question

Match the properties of the species in Column I with the calculated number in Column II.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ i.& \text{Number of protons in the element with mass number 81}& p.& 50\\ & \text{and having}31.7\%\text{more neutrons than protons}& & \\ ii.& \text{Ionic mass of}{M}^{2+}\text{which is isoelectronic of}C{O}_2\text{and has}& q.& 26\\ & \text{(Z+2) neutrons}.& & \\ iii.& \text{Ionic mass of}{X}^{-}\text{with 17 protons and having}11.1\%& r.& 35\\ & \text{more neutrons than electrons}& & \\ iv.& \text{Atomic number of}{M}^{3+}\text{with mass number 56 and having}& s.& 37\\ & 30.4\%\text{more neutrons than electrons}& & \end{array}$

• A. $\begin{array}{cc}iii& iv\\ p& q\end{array}$
• B. $\begin{array}{cc}i& ii\\ q& p\end{array}$
• C. $\begin{array}{cc}iii& iv\\ s& q\end{array}$
• D. $\begin{array}{ccc}i& ii& \\ r& q& \end{array}$

Answer 1

Answer: (C), (D)

The correct options are
C $\begin{array}{cc}iii& iv\\ s& q\end{array}$
D $\begin{array}{ccc}i& ii& \\ r& q& \end{array}$

i) $N+P=81$
$N=31.7\%$ more than protons
Then, $N=(P+0.317P)$
$\therefore (P+0.317P+P)=81$
$2.317P=81$
$P=35$
Thus, $\left(i\right)\to \left(r\right)$

ii) $M^{2+}$ has $=6+16=22\text{electrons}$
$M^{2+}$ has $=24\text{protons}$
$Z\left(M^{2+}\right)=24$
$N=24+2$
$=26$
Therefore, ionic mass $=24+26=50$
Thus, $\left(ii\right)\to \left(q\right)$

iii) $X^{-}$ has protons $=17$
Electrons $=17+1=18$
From the data, number of neutrons is equal to the sum of number of electrons and 11.1% of number of electrons.
Neutrons $=(18+\frac{18\times 11.1}{100})$
$=18+2$
$=20$
Therefore, ionic mass of $X=P+N=37$
Thus, $\left(iii\right)\to \left(s\right)$

iv) $N+P=56$
Electrons in $M^{3+}=(P-3)$
Total number of neutrons, from the data given, is equal to the number of electrons and 30.4% of number of electrons
$N=(P-3)+(P-3)\times \frac{30.4}{100}$
$N=(P-3)+0.304P-0.912=56$
$P=26$
Thus, $\left(iv\right)\to \left(q\right)$