NEET Practice Questions (MCQs) with Answers & Solutions

(c) The electronic configuration  1s²2s²2p⁵3s¹ shows lowest ionisation energy because this configuration is unstable due to the presence of one electron in s-orbital. Hence, less energy is required to remove the electron.

A cation has always the lesser ionic size than a metal atom due to loss of electron and an anion has always the greater size than metal atom due to gain of electrons. So, the correct order of size is as

          Ca²⁺<K⁺<Ar<Cl<S^2-

In NH3, the electronegativity difference between N and H causes the electron cloud to be displaced towards the more electronegative H atoms. This results in a bond dipole moment in the same direction as the atomic dipole moment of N. In NF3, the highly electronegative F atoms displace the electron cloud away from N, causing the bond dipole moments to be in the opposite direction to the atomic dipole moment of N. This cancellation effect in NF3 results in a smaller net dipole moment compared to NH3.

(d) The correct order of the mobility of the alkali metal ions in aqueous solutions. Rb⁺ > K⁺ > Na⁺ > Li⁺ due to following order of hydration energy of these ions Li⁺ > Na⁺ > K⁺ > Rb⁺ and due to hydration of ion, mobility decreases.

(b) Ununtrium for atomic number = 113;  for Z>86 pd.no. =7

     $$\begin{gathered} Ununtrium = {\left[\mathrm{Rn}\right]}^{86} 7s^{2 }, 5f^{14},6d^{10 },7p^1 \\ p‐block \end{gathered}$$

nihonium is  [Rn].5f¹⁴.6d¹⁰.7s².7p¹

(b) Tantalum (Z= 73) is a transition element.

(c)    Period number     Group number      Element

             5                            7                      Tc

             6                           13                      Tl

             6                            5                       Ta

$$\begin{gathered} \left[Xe\right]4f^{14}5d^{1}6s^2 \\ Atomic number =71 Period number =6th Group number = 3rd \end{gathered}$$

(b) Rubidium is most electropositive in nature.

(c)  $$\begin{gathered} \left(a\right) Cr : 4s^{1}3d^s (each d‐orbital has le") \\ \left(b\right) {\mathrm{Fe}}^{3+}: 4{\mathrm{s}}^{0}3{\mathrm{d}}^{\mathrm{s} } (\mathrm{each} \mathrm{d}‐\mathrm{orbital} \mathrm{has} \mathrm{le}") \\ \left(\mathrm{c}\right) {\mathrm{Cu}}^{+}:4{\mathrm{s}}^{0}3{\mathrm{d}}^{10} (\mathrm{each} \mathrm{d}‐\mathrm{orbital} \mathrm{has} 2\mathrm{e}") \end{gathered}$$