The law states that when lighter elements are arranged in order of their increasing atomic weights, the properties of every eighth element were similar to those of the first one like the eighth note of a musical scale.
2.
When we go from left to right in a period,
A.
the basic nature of the oxides increases
B.
the basic nature of the oxides decreases
C.
there is no regular trend in the nature of oxides
D.
oxides of only last two groups are basic in nature
The basic nature of the oxides decreases from left to right in a period. e.g., in third period,
$$\mathop {N{a_2}O}\limits_{{\text{Strongly basic}}} > \mathop {MgO}\limits_{{\text{Basic}}} > \mathop {A{l_2}{O_3}}\limits_{{\text{Amphoteric}}} > $$ $$\mathop {Si{O_2}}\limits_{{\text{Weakly acidic}}} > \mathop {{P_4}{O_{10}}}\limits_{{\text{Acidic}}} > \mathop {S{O_2}}\limits_{{\text{Strongly}}\,\,{\text{acidic}}} $$ $$ > \mathop {C{l_2}{O_7}}\limits_{{\text{Strongly}}\,\,{\text{acidic}}} $$
3.
The first ionisation potential $$\left( {{\text{in}}\,\,eV} \right)$$ of $$Be$$ and $$B,$$ respectively are
First ionisation potential of beryllium $$(Be)$$ is greater than boron $$(B)$$ due to stable configuration
$${}_4Be = 1{s^2},2{s^2}$$
$${}_5B = 1{s^2},2{s^2}2{p^1}$$
Order of attraction of electrons towards nucleus is $$2s > 2p,$$ so more amount of energy is required to remove the electron from $$2s$$ orbital in comparison to $$2p$$ orbital. So, ionisation potential of $$Be$$ is 9.32 $$eV$$ and $$B$$ is 8.29 $$eV.$$
4.
According to the law of triads,
A.
the properties of the middle element were in between those of the other two members
B.
three elements arranged according to increasing weights have similar properties
C.
the elements can be grouped in the groups of six elements
D.
every third element resembles the first element in periodic table
Answer :
the properties of the middle element were in between those of the other two members
Official name of an element with atomic number 117 is Tennessine $$(Ts)$$ given by IUPAC.
6.
Sum of first three ionization energies of $$Al$$ is $$53.0\,eV\,ato{m^{ - 1}}$$ and the sum of first two ionization energies of $$Na$$ is $$52.2\,eV\,ato{m^{ - 1}}.$$ Out of $$Al\left( {{\text{III}}} \right)$$ and $$Na\left( {{\text{II}}} \right)$$
A.
$$Na\left( {{\text{II}}} \right)$$ is more stable than $$Al\left( {{\text{III}}} \right)$$
B.
$$Al\left( {{\text{III}}} \right)$$ is more stable than $$Na\left( {{\text{II}}} \right)$$
C.
Both are equally stable
D.
Both are equally unstable
Answer :
$$Al\left( {{\text{III}}} \right)$$ is more stable than $$Na\left( {{\text{II}}} \right)$$
An atom has electronic configuration
$$1{s^2},2{s^2},2{p^6}\,3{s^2}3{p^6}3{d^3},4{s^2}$$
It is a member of $$d$$- block element because the last electron is filled in $$d$$- subshell as $$3{d^3}$$ and the following electronic configuration is possible for $$d$$-subshell as $$\left( {n - 1} \right){d^{\left( {1\,to\,10} \right)}}$$
\[\underset{\begin{smallmatrix}
\\
\\
n{{s}^{2}}\left( n-1 \right){{s}^{2}}{{p}^{6}}
\end{smallmatrix}}{\mathop{\text{Group}\,\,\text{number}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
3 \\
{{d}^{1}}
\end{smallmatrix}}{\mathop{\text{III}\,B}}\,\,\,\underset{\begin{smallmatrix}
4 \\
{{d}^{2}}
\end{smallmatrix}}{\mathop{\,\,\,\text{IV}B}}\,\] \[\underset{\begin{smallmatrix}
5 \\
{{d}^{3}}
\end{smallmatrix}}{\mathop{\text{V}B}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
6 \\
{{d}^{4}}
\end{smallmatrix}}{\mathop{\text{VI}\,B}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
7 \\
{{d}^{5}}
\end{smallmatrix}}{\mathop{\text{VII}\,B\,\,}}\,\,\,\,\underset{\begin{smallmatrix}
8 \\
{{d}^{6}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\] \[\underset{\begin{smallmatrix}
9 \\
{{d}^{7}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
10 \\
{{d}^{8}}
\end{smallmatrix}}{\mathop{\text{VIII}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
11 \\
{{d}^{9}}
\end{smallmatrix}}{\mathop{\text{IB}}}\,\,\,\,\,\,\underset{\begin{smallmatrix}
12 \\
{{d}^{10}}
\end{smallmatrix}}{\mathop{\text{IIB}}}\,\]
Hence, it is member of third group