MCQ
Last element of group-$IV$ is found to be
- AStrong metallic
- ✓Weak metallic
- CStrong non-metallic
- DWeak non-metallic
✓
Answer
Correct option: B.
Weak metallic
b
Down the group metallic character increases. Group $IV$ shows variation from non-metallic character to weak metallic character as we move down the group.
Down the group metallic character increases. Group $IV$ shows variation from non-metallic character to weak metallic character as we move down the group.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Arrange the following alkanes in decreasing order of their heats of combustion
→$(i)\,\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,C{H_3}}
\end{array}}\limits_{(Neo - pen\tan e)\,(i)} $
$(ii\mathop {)\,\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,\,\,\,\,C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}}\limits_{(Iso - pen\tan e)\,\,(ii)} $
$(iii)\,\mathop {C{H_3} - C{H_2} - C{H_2} - C{H_2} - C{H_3}}\limits_{(n - pen\tan e)} $
How many moles of $P_4$ can be produced by reaction of $0.10$ moles $Ca_5(PO_4)_3F, 0.36$ moles $SiO_2$ and $0.90$ moles $C$ according to the following reaction?
→$4C{a_5}{(P{O_4})_3}F + 18Si{O_2} + 30C \to 3{P_4} + 2Ca{F_2} + 18CaSi{O_3} + 30CO$
Which pair will form the most stable ionic bond?
Lead is considered as
The product $A$ formed in the following reaction is: $Image$
→In an element going away from nucleus, the energy of particle
Which of the following is an chiral compound
The single, double and triple bond lengths of carbon in carbon dioxide are respectively
$Ph - MgCl\xrightarrow{{Hydrolysis}}$ Product
→Product will be
Among the following compounds, the increasing order of their basic strength is
