Download App

-p-Block elements - I — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry-p-Block elements - I1 Mark
MCQ
Boron does not form $B^{3+}$ cation easily. it is due to
  • ✓
    energy required to form $B^{3+}$ ion is far more than that which would be compensated by lattice energies or hydration energies of such ion
  • B
    boron is non-metal
  • C
    boron is semiconductor
  • D
    none of the above
✓

Answer

Correct option: A.
energy required to form $B^{3+}$ ion is far more than that which would be compensated by lattice energies or hydration energies of such ion
a
As the Boron atom is small in size a large amount of energy is needed to remove $3$ electrons from the boron atom. So Boron does not form $B_3+$ ion. The atomic number of Boron is $5$ . Its electronic configuration is $1s [2] 2 s[2] 2 p[1]$

When one electron is removed from the $p$ orbital a $He$-like fulfilled s orbital is left. This is highly stable. So the second ionization enthalpy is quite high. Again when one electron is removed a half filled orbital is left. So the third ionization enthalpy is also quite high.

Since the total energy needed to make $B _3+$ is the total of all the ionization enthalpies an enormous amount of energy is required to form it.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from -p-Block elements - I→-p-Block elements - I — all question types→Chemistry STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Match list $-I$ with list $-II$ and then select the correct answer from the codes given below that list

                                  List $-I$

$(A){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_6} + CO + HCl\,\xrightarrow{{AlC{l_3}}}\,{C_6}{H_5}CHO$

$(B){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_5} + C{H_3}$ $\xrightarrow{{Cr{O_2}C{l_2}}}\,{C_6}{H_5}CHO$

$(C){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_5}C{H_2}Br\, + \,C{H_3}Br$ $\xrightarrow{{Na/ether}}\,{C_6}{H_5}C{H_2}C{H_3}$

$(D){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {C_6}{H_6}\, + \,{(C{H_3})_2}C = C{H_2}$ $\xrightarrow{{{H_2}S{O_4}}}\,{C_6}{H_5}C{(C{H_3})_3}$

                                 List $-II$
$(a)$   Friedel-Crafts reaction
$(b)$   Wurtz-Fitting reaction
$(c)$   Gattermann-Koch Aldehyde synthesis
$(d)$   Etard’s reaction

→
Aqueous solution of two compounds $M_1OH$ and $M_2OH$ are prepared in two different beakers if the $EN$ of $M_1 = 3.4, EN$ of $M_2 = 1.2, EN$ of $O = 3.5$ and $EN$ of $H = 2.1$, then nature of two solutions will be
→
The following compounds differ in respect of 
→
The reaction of toluene with $Cl_2$ in presence of $FeCl_3$ gives $X$ and reaction in presence of light gives $Y.$ Thus, $X$ and $Y$ are
→
The number of sodium atoms in $2 $ moles of sodium ferrocyanide is
→
Which of the following sets of orbitals may degenerate
→
When two atoms of chlorine combine to form one molecule of chlorine gas, the energy of the molecule
→
$750$ waves of a energy radiation passes in one minute from a point then what will be wavelength of radiation
→
For $3d_{z^2}$ calculate the value of $n, \ell , m$ and $s$
→
$\begin{array}{*{20}{c}}
  {2C{H_3} - C = C{H_2}} \\ 
  {\,|} \\ 
  {\,\,\,Ph} 
\end{array}\xrightarrow{{H - A\,(acid)}}\mathop {(A)}\limits_{(major)} $

Product $(A)$ is

→