
There are four elements ‘$p$’, ‘$q$’, ‘$r$’ and ‘$s$’ having atomic numbers $Z - 1,Z,Z + 1$ and $Z + 2$ respectively. If the element ‘$q$’ is an inert gas, select the correct answers from the following statements.
A) ‘$p$’ has most negative electron gain enthalpy in the respective period.
B) ‘$r$’ is an alkali metal
C) ‘$s$’ exists in $ + 2$ oxidation state.
Answer
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Hint: Know the position of groups and their properties well. The change in different properties on moving down the group and moving left to right in a period should be considered. The common names of certain groups like Group one – alkali metals, Group 2- alkaline earth metals, etc. should be noted.
Complete answer:
It is given that element ‘$q$’ is an inert gas having atomic number$Z$.
As we know that electrons gain enthalpy of elements present just before the inert gas group have the highest enthalpy as they want to gain an electron and convert themselves into stable and inert state. For e.g. in the periodic table, elements of group seventeen (halogens) have highest electron gain enthalpy.
$\therefore $Element ‘$p$’ having atomic number $Z - 1$ has the most negative (very high) electron gain enthalpy.
In the periodic table, after the inert gas family (group 18), comes the group of alkali metals (group 1). Thus the element ‘$r$’ having atomic number $Z + 1$ belongs to the alkali metal family.
After group 1 i.e., alkali metals came the group of alkaline earth metals (group 2). Elements in this group exist mostly in $ + 2$ oxidation state. For e.g. Magnesium ( $M{g^{2 + }}$), Calcium ( $C{a^{2 + }}$) etc.
$\therefore $Element ‘$s$’ having atomic number $Z + 2$ belongs to alkaline earth metals and exists in $ + 2$ oxidation state.
Hence all options are correct.
Note: Properties of respective groups should be prepared well. The particular stable oxidation states of respective groups of periodic tables must be considered as they will be used in answering these questions. Elements of group one exist in $ + 1$ oxidation state and elements of group 2 exist in $ + 2$ oxidation state.
Complete answer:
It is given that element ‘$q$’ is an inert gas having atomic number$Z$.
As we know that electrons gain enthalpy of elements present just before the inert gas group have the highest enthalpy as they want to gain an electron and convert themselves into stable and inert state. For e.g. in the periodic table, elements of group seventeen (halogens) have highest electron gain enthalpy.
$\therefore $Element ‘$p$’ having atomic number $Z - 1$ has the most negative (very high) electron gain enthalpy.
In the periodic table, after the inert gas family (group 18), comes the group of alkali metals (group 1). Thus the element ‘$r$’ having atomic number $Z + 1$ belongs to the alkali metal family.
After group 1 i.e., alkali metals came the group of alkaline earth metals (group 2). Elements in this group exist mostly in $ + 2$ oxidation state. For e.g. Magnesium ( $M{g^{2 + }}$), Calcium ( $C{a^{2 + }}$) etc.
$\therefore $Element ‘$s$’ having atomic number $Z + 2$ belongs to alkaline earth metals and exists in $ + 2$ oxidation state.
Hence all options are correct.
Note: Properties of respective groups should be prepared well. The particular stable oxidation states of respective groups of periodic tables must be considered as they will be used in answering these questions. Elements of group one exist in $ + 1$ oxidation state and elements of group 2 exist in $ + 2$ oxidation state.
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