Valency and oxidation state Questions in English

(A) The formation of a cation occurs by the loss of electrons from the valence shell of an atom.
For the transition of $Zn$ $([Ar] 3d^{10} 4s^2)$ to $Zn^{2+}$ $([Ar] 3d^{10})$,two electrons are removed from the $4s$ orbital.
Therefore,there is a decrease in the number of valence electrons.

(C) The correct option is $C$. $6$.
In the structure of sulphuric acid $(H_2SO_4)$,the sulphur atom is bonded to four oxygen atoms.
Two oxygen atoms are bonded via double bonds $(S=O)$ and two oxygen atoms are bonded via single bonds $(S-OH)$.
Counting the total number of bonds formed by the sulphur atom,we get $2 + 2 + 1 + 1 = 6$ bonds.
Therefore,the valency of sulphur in $H_2SO_4$ is $6$.

(B) To find the oxidation state of phosphorus $(P)$ in $H_3PO_4$,let its oxidation state be $x$.
The oxidation states of $H$ and $O$ are $+1$ and $-2$ respectively.
The sum of oxidation states in a neutral molecule is $0$.
$3(+1) + x + 4(-2) = 0$
$3 + x - 8 = 0$
$x - 5 = 0$
$x = +5$.
Thus,the oxidation state of phosphorus is $5$.

(D) The element $M$ has $3$ valence electrons,so its valency is $3$ and it forms $M^{3+}$ ions.
The element $A$ has $7$ valence electrons,so its valency is $1$ and it forms $A^{-}$ ions.
To form a neutral compound,one $M^{3+}$ ion combines with three $A^{-}$ ions.
Therefore,the chemical formula of the compound is $MA_3$.

(A) According to the criss-cross rule,the valency of $A$ $(3)$ becomes the subscript of $B$,and the valency of $B$ $(2)$ becomes the subscript of $A$.
Thus,the formula of the compound is $A_2B_3$.

(A) In acetylene $(C_2H_2)$,the structure is $H-C \equiv C-H$.
Each carbon atom forms one single bond with a hydrogen atom and a triple bond with the other carbon atom.
Since each carbon atom is involved in $4$ covalent bonds,its valency is $4$.

(D) The acidity of oxoacids is generally proportional to the oxidation state of the central atom.
Calculating the oxidation states:
For $H_2SO_3$: $2(+1) + x + 3(-2) = 0 \implies x = +4$.
For $H_3PO_3$: $3(+1) + x + 3(-2) = 0 \implies x = +3$.
For $HClO_3$: $1(+1) + x + 3(-2) = 0 \implies x = +5$.
Comparing the oxidation states: $Cl (+5) > S (+4) > P (+3)$.
Therefore,the decreasing order of acidity is $III > I > II$.

(A) The electronic configuration of phosphorus $(P)$ is $[Ne] 3s^2 3p^3$.
It can gain $3$ electrons to achieve a stable noble gas configuration ($-3$ state) or lose up to $5$ electrons (valence shell electrons) to reach a $+5$ state.
Thus,the oxidation states of phosphorus range from $-3$ to $+5$.

(A) The element with atomic number $13$ is Aluminum $(Al)$.
Its electronic configuration is $[Ne] 3s^2 3p^1$.
It loses $3$ electrons to achieve a stable noble gas configuration,thus it shows a $+3$ oxidation state.

(C) The most common oxidation state of $-2$ is typically exhibited by elements of group $16$ (chalcogens),such as oxygen and sulfur.
These elements have an electronic configuration of $ns^2 np^4$ in their outermost shell.
Therefore,the number of electrons present in the outermost shell is $2 + 4 = 6$.

(C) The oxidation states of the given elements are as follows:
$N$: $-3$ to $+5$
$O$: $-2$ to $+2$
$Cl$: $-1$ to $+7$
$C$: $-4$ to $+4$
Comparing these,$Cl$ exhibits the highest oxidation number of $+7$.

(B) Let the oxidation state of $I$ be $x$.
In the phosphate ion,$PO_4^{3-}$,the oxidation state of the group is $-3$.
Since the compound $IPO_4$ is neutral,the sum of the oxidation states must be zero:
$x + (-3) = 0$
$x = +3$
Therefore,the oxidation state of $I$ is $+3$.

(C) Valence electrons: Each group is characterized by its valence electrons,i.e.,the number of electrons in the outermost shell of the elements are the same within a group. Thus,the common oxidation states for the elements of the same group are fixed.
Due to the same number of valence electrons,elements of the same group show similar chemical properties.
Atomic size: Moving down the group,a new shell is added with each period.
The electron in the new shell also increases shielding. Therefore,the $Z_{eff}$ decreases along with which the attractive force of the nucleus decreases,and thus the size increases.
Density: Density $= \frac{\text{Mass}}{\text{Volume}}$. Moving down the group,mass as well as volume increases. Therefore,density depends on the dominating factor. Usually,down the group,density increases,i.e.,the effect of mass dominates.
Metallic character: Moving down the group,the nuclear attractive force on outer electrons decreases due to a decrease in $Z_{eff}$ which is due to an increase in shielding by inner shells. As a result,the ionization energy decreases down the group,and therefore,metallic character increases.
Hence,the correct answer is valence electrons.

(C) The valency of an element with respect to oxygen is defined as the number of oxygen atoms that combine with one atom of the element.
For elements in a period,the valency with respect to oxygen increases from $1$ to $4$ as we move from group $IA$ to $IVA$.
After group $IVA$,the valency with respect to oxygen decreases from $3$ to $1$ as we move from group $VA$ to $VIIA$ (e.g.,$Na_2O$ (valency $1$),$MgO$ (valency $2$),$Al_2O_3$ (valency $3$),$SiO_2$ (valency $4$),$P_2O_5$ (valency $5$ is oxidation state,but valency w.r.t oxygen is $3$ for $P_2O_3$ or $5$ for $P_2O_5$ depending on the oxide,generally it follows the pattern $1, 2, 3, 4, 3, 2, 1$)).
Thus,the valency increases from $IA$ to $IVA$ and then decreases from $VA$ to $VIIA$.

(B) The element with atomic number $7$ is Nitrogen $(N)$.
The electronic configuration of Nitrogen is $1s^2, 2s^2, 2p^3$.
The valence shell is the second shell $(n=2)$,which contains $2+3 = 5$ electrons.
While Nitrogen typically exhibits a valency of $3$ due to its $3$ unpaired $p$-electrons,the maximum covalency it can show is $4$ (e.g.,in $NH_4^+$) because it lacks $d$-orbitals.
However,in the context of valence electrons,the number of valence electrons is $5$.

(C) Elements in group $III$ (or group $13$) typically exhibit a valency of $3$ due to their valence shell configuration of $ns^2 np^1$.
When these elements react with oxygen,they form oxides with the general formula $R_2O_3$,where $R$ is in the $+3$ oxidation state.
For example,Aluminum $(Al)$ belongs to group $III$ and forms Aluminum oxide $(Al_2O_3)$.
Therefore,the correct statement is that it forms $R_2O_3$.

(A) Lead $(Pb)$ belongs to group $14$ of the periodic table.
It exhibits two common oxidation states: $+2$ and $+4$.
The $+2$ oxidation state is more stable than the $+4$ oxidation state due to the inert pair effect,where the $6s^2$ electrons are reluctant to participate in bonding.

(C) To find the oxidation state of phosphorus $(x)$ in $H_3PO_2$:
$3(+1) + x + 2(-2) = 0$
$3 + x - 4 = 0$
$x - 1 = 0$
$x = +1$
Therefore,the oxidation state of phosphorus in $H_3PO_2$ is $+1$.

(A) Fluorine is the only element that shows only one oxidation state,i.e.,$-1$.
This is because Fluorine is the most electronegative element and lacks $d$-orbitals in its valence shell.
In contrast,Chlorine,Bromine,and Iodine exhibit multiple oxidation states such as $-1, +1, +3, +5$,and $+7$ due to the presence of vacant $d$-orbitals.
Therefore,the correct answer is $F$.

(C) Inert gases are unreactive due to their complete valence shell configuration (octet or duplet).
Therefore,the valency of inert gases is $0$.

(B) The correct option is $(B)$.
$Cu$ (Copper) shows variable valency (e.g.,$+1$ and $+2$) due to the presence of vacant $d-$orbitals in its electronic configuration,which allows for the participation of $d-$electrons in bonding.

(C) The valencies of the given elements are as follows:
$Al$ (Aluminum) has a valency of $3$.
$Rb$ (Rubidium) has a valency of $1$.
$Cu$ (Copper) commonly exhibits a valency of $2$ (as in $Cu^{2+}$).
$Ge$ (Germanium) has a valency of $4$.

(C) The valency of elements in a period increases one by one from $I \, A$ to $IV \, A$ and then decreases one by one from $V \, A$ to $VII \, A$.

(B) The electronic configuration $1s^2 2s^2 2p^6 3s^1$ indicates that the metal $M$ has $1$ valence electron in its outermost shell.
Therefore,the valency of the metal $M$ is $+1$.
Oxygen typically forms an oxide ion with a valency of $-2$ $(O^{2-})$.
By applying the criss-cross method for the valencies of $M^{+1}$ and $O^{-2}$,the chemical formula of the oxide is $M_2O$.

(B) The valency of $H$ is typically $+1$ or $-1$.
The valency of $Na$ is constant at $+1$.
The valency of $Fe$ is variable,being $+2$ or $+3$.
The valency of $O$ can vary (e.g.,$-2, -1, +1, -\frac{1}{2}$).

(A) The electronic configuration of $X$ is $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^2$. It has $2$ electrons in its outermost shell,so its valency is $+2$.
The electronic configuration of $Y$ is $1s^2, 2s^2 2p^6, 3s^2 3p^5$. It has $7$ electrons in its outermost shell,so it needs $1$ electron to complete its octet,making its valency $-1$.
To form a neutral compound,the charges must balance. Thus,one atom of $X$ $(+2)$ combines with two atoms of $Y$ ($-1$ each) to form $XY_2$.

(B) The inert pair effect is the tendency of the $ns^2$ electron pair to remain unionized due to the poor shielding of the intervening $d$ and $f$ orbitals,which increases as we move down a group.
In group $14$,the stability of the $+4$ oxidation state decreases down the group,while the stability of the $+2$ oxidation state increases.
For $Sn$ ($Group$ $14$),the $+4$ oxidation state is more stable than the $+2$ oxidation state because the inert pair effect is less pronounced in $Sn$ compared to $Pb$.
Thus,$Sn^{+4} > Sn^{+2}$ is the correct order of stability.
In contrast,$Pb^{+2} > Pb^{+4}$,$Bi^{+3} > Bi^{+5}$,and $PbCl_4 > PbBr_4$ (due to the oxidizing nature of $Pb^{+4}$ and the reducing nature of halides).

(A) Copper pyrites is represented as $CuFeS_2$.
In this compound,the oxidation state of copper is $+1$,the oxidation state of iron is $+2$,and the oxidation state of sulfur is $-2$.
Thus,the oxidation state of iron in $CuFeS_2$ is $+2$.

(B) To determine the correct formula,we analyze the valency of element $X$ in each compound.
$X_2O_3$ implies $X$ has a valency of $+3$ $(X^{3+})$ and $O$ has a valency of $-2$ $(O^{2-})$,which is chemically valid.
$X_2Cl_3$ implies $X$ has a valency of $+3$ $(X^{3+})$ and $Cl$ has a valency of $-2$ $(Cl^{2-})$. Since the chloride ion is $Cl^-$,this formula is incorrect.
$X_2(SO_4)_3$ implies $X^{3+}$ and $SO_4^{2-}$,which is chemically valid.
$XPO_4$ implies $X^{3+}$ and $PO_4^{3-}$,which is chemically valid.
Therefore,$X_2Cl_3$ is the incorrect formula. The correct formula for the chloride of $X$ would be $XCl_3$.

(A) The electronic configuration of an element with atomic number $15$ is $1s^2, 2s^2, 2p^6, 3s^2, 3p^3$.
Since the valence shell is the $3rd$ shell $(n=3)$,the number of valence electrons is $2 + 3 = 5$.
For $p$-block elements,the group number is calculated as $10 + \text{valence electrons} = 10 + 5 = 15$.
The valency is calculated as $8 - \text{valence electrons} = 8 - 5 = 3$.
Thus,the group number is $15$,the number of valence electrons is $5$,and the valency is $3$.

(D) The ionization energy values are $IE_1 = 100 \ kJ/mol$,$IE_2 = 250 \ kJ/mol$,and $IE_3 = 1500 \ kJ/mol$.
There is a large jump between $IE_2$ and $IE_3$,which indicates that the element $M$ has $2$ valence electrons.
Since the element has $2$ valence electrons,its valency is $2$.
Therefore,the formula of its oxide is $MO$.

(C) The atomic size of $Iodine$ is large and it possesses empty $d$-orbitals,allowing for the expansion of its valence shell.
In its $3^{rd}$ excited state,$Iodine$ has $7$ unpaired electrons,enabling it to exhibit positive oxidation states when bonded to more electronegative elements like $F$,$Cl$,or $Br$.
Generally,all halogens exhibit a negative oxidation state of $-1$ due to their high electronegativity,but $Iodine$ is unique among them in its ability to readily show positive oxidation states as well.

(A) The atomic number of an element is equal to the number of protons in its nucleus. Since the element has $9$ protons,its atomic number is $9$.
The electronic configuration of this element is $2, 7$.
To achieve a stable octet (noble gas configuration),the element needs to gain $1$ electron.
Therefore,the valency of the element is $1$.

(C) Element $P$ belongs to Group $15$ (Nitrogen family),so its valency is $3$ (it gains $3$ electrons to complete its octet).
Element $Q$ belongs to Group $2$ (Alkaline earth metals),so its valency is $2$ (it loses $2$ electrons to achieve noble gas configuration).
To form a neutral compound,the total positive charge must equal the total negative charge.
Using the criss-cross method: $Q^{2+}$ and $P^{3-}$ combine to form $Q_3P_2$.
However,based on standard chemical nomenclature where the metal (cation) is written first,the formula is $Q_3P_2$. Given the options provided,the ratio of $P$ to $Q$ is $2:3$,which corresponds to $P_2Q_3$.

(B) To find the formula,we need to balance the total positive and negative charges.
Let the formula be $A_x(BC_y)_z$.
The oxidation state of $A$ is $+2$.
The oxidation state of the polyatomic group $(BC_y)$ is calculated as $B + y \times C = +5 + y(-2) = 5 - 2y$.
For the compound to be neutral,the total charge must be zero: $x(+2) + z(5 - 2y) = 0$.
Testing option $B$: $A_3(BC_4)_2$.
Here,$x=3$,$y=4$,$z=2$.
Charge of $A = 3 \times (+2) = +6$.
Charge of $(BC_4) = +5 + 4(-2) = 5 - 8 = -3$.
Total charge = $3(+2) + 2(-3) = +6 - 6 = 0$.
Thus,the formula is $A_3(BC_4)_2$.

(C) For $ClO_n^-$ ions,as $n$ increases from $1$ to $4$:
$1$. Oxidation number of $Cl$ increases: $ClO^- (+1), ClO_2^- (+3), ClO_3^- (+5), ClO_4^- (+7)$.
$2$. Bond order increases: $ClO^- (1.0), ClO_2^- (1.5), ClO_3^- (1.67), ClO_4^- (1.75)$.
$3$. Bond length decreases as bond order increases.
$4$. Hybridization of $Cl$ remains $sp^3$ in all cases.
Therefore,the correct statement is that on increasing the value of $n$,the oxidation number of the central atom increases.

(D) In cyclic sulfur trioxide,which is the trimer $(SO_3)_3$ or $S_3O_9$,the sulfur atom is bonded to oxygen atoms.
Using the oxidation state rules,let the oxidation state of sulfur be $x$.
For the molecule $(SO_3)_3$,the total charge is $0$.
$3 \times [x + 3 \times (-2)] = 0$
$3 \times [x - 6] = 0$
$x - 6 = 0$
$x = +6$.
Thus,the oxidation state of sulfur is $+6$.

(B) Fluorine $(F)$ is the most electronegative element and only shows a $-1$ oxidation state.
Iodine $(I)$ is a halogen that exhibits both negative $(-1)$ and positive $(+1, +3, +5, +7)$ oxidation states due to the presence of vacant $d$-orbitals.
Lithium $(Li)$ is an alkali metal and only shows a $+1$ oxidation state.
Helium $(He)$ is a noble gas and is chemically inert,showing an oxidation state of $0$.

(C) Sodium $(Na)$ and Potassium $(K)$ are highly electropositive elements and typically exhibit a fixed oxidation state of $+1$.
Fluorine $(F)$ is the most electronegative element and exhibits only a $-1$ oxidation state.
Chlorine $(Cl)$ exhibits variable oxidation states ranging from $-1$ to $+7$ (e.g.,$-1, +1, +3, +5, +7$) due to the presence of vacant $d$-orbitals.

(D) Sulphur belongs to group $16$ of the periodic table and has the valence shell electronic configuration $ns^2 np^4$.
It can exhibit a range of oxidation states due to the presence of vacant $d$-orbitals.
Common oxidation states shown by sulphur include $-2$ (as in $H_2S$),$+2$ (as in $SCl_2$),$+4$ (as in $SO_2$),and $+6$ (as in $SO_3$ or $H_2SO_4$).

(B) $P$ is a Group $2$ element,which has $2$ valence electrons,so its valency is $2$.
$Q$ is a Group $15$ element,which has $5$ valence electrons,so it needs $3$ electrons to complete its octet,making its valency $3$.
To form a neutral compound,the charges must be balanced: $P^{2+}$ and $Q^{3-}$.
By crossing the valencies,the formula becomes $P_3Q_2$.

(C) The given electronic configuration is $ns^2 np^1$.
This indicates that the element $(A)$ has $3$ valence electrons.
Therefore,the valency of element $(A)$ is $+3$.
The valency of oxygen $(O)$ is $-2$.
By crossing the valencies,the formula of the oxide is $A_2O_3$.

(D) Valency is defined as the combining capacity of an element.
Noble gases,such as $Kr$ (Krypton),have a completely filled valence shell ($ns^2 np^6$ configuration).
Due to their stable electronic configuration,they have very little tendency to gain,lose,or share electrons,resulting in a valency of zero.

(A) In an elemental form like $S_8$,the oxidation state of an atom is always $0$ because the atoms are bonded to identical atoms.
In the $S_8$ molecule,each sulfur atom is bonded to two other sulfur atoms by single covalent bonds,so its covalency is $2$.

(A) The oxidation state of an element in its compounds is determined by its group number and valence electrons.
$Cs$ (Cesium) belongs to Group $1$ and exhibits an oxidation state of $+1$.
$Ca$ (Calcium) belongs to Group $2$ and exhibits an oxidation state of $+2$.
$Mg$ (Magnesium) belongs to Group $2$ and exhibits an oxidation state of $+2$.
$Al$ (Aluminum) belongs to Group $13$ and exhibits an oxidation state of $+3$.
Comparing these values,$Cs$ has the lowest oxidation state of $+1$.

(B) The Assertion is correct: $HClO_4$ is indeed a stronger acid than $HClO_3$.
The Reason is also correct: The oxidation state of $Cl$ in $HClO_4$ is $+VII$ and in $HClO_3$ is $+V$.
However,the Reason is not the correct explanation for the Assertion. The acidity of oxyacids is determined by the number of non-hydroxylated oxygen atoms $(n)$ in the general formula $(HO)_m ZO_n$.
For $HClO_4$ $(HOClO_3)$,$n = 3$. For $HClO_3$ $(HOClO_2)$,$n = 2$.
As the number of non-hydroxylated oxygen atoms $(n)$ increases,the electron-withdrawing effect on the $Cl$ atom increases,which in turn weakens the $O-H$ bond,facilitating the release of the $H^+$ ion. Thus,the acidity depends on the number of non-hydroxylated oxygen atoms,not directly on the oxidation state of the central atom.

(N/A) Silicon is a group $14$ element with a valency of $4$; bromine belongs to the halogen family (group $17$) with a valency of $1$. By cross-multiplying the valencies,the formula of the compound formed is $SiBr_4$.
$(b)$ Aluminium belongs to group $13$ with a valency of $3$; sulphur belongs to group $16$ with a valency of $2$. By cross-multiplying the valencies,the formula of the compound formed is $Al_2S_3$.

(N/A) Lithium $(Li)$ has a valency of $1$ and oxygen $(O)$ has a valency of $2$. The formula is $Li_2O$.
$(b)$ Magnesium $(Mg)$ has a valency of $2$ and nitrogen $(N)$ has a valency of $3$. The formula is $Mg_3N_2$.
$(c)$ Aluminium $(Al)$ has a valency of $3$ and iodine $(I)$ has a valency of $1$. The formula is $AlI_3$.
$(d)$ Silicon $(Si)$ has a valency of $4$ and oxygen $(O)$ has a valency of $2$. The formula is $SiO_2$.
$(e)$ Phosphorus $(P)$ can show valencies of $3$ or $5$ and fluorine $(F)$ has a valency of $1$. The formulas are $PF_3$ or $PF_5$.
$(f)$ The element with atomic number $71$ is Lutetium $(Lu)$,which has a valency of $3$. Fluorine $(F)$ has a valency of $1$. The formula is $LuF_3$.

To represent compounds using Stock notation,we determine the oxidation state of the metal and write it in Roman numerals in parentheses after the metal symbol.
$1$. $HAuCl_{4}$: Oxidation state of $Au$ is $+3$,so it is $HAu(III)Cl_{4}$.
$2$. $Tl_{2}O$: Oxidation state of $Tl$ is $+1$,so it is $Tl_{2}(I)O$.
$3$. $FeO$: Oxidation state of $Fe$ is $+2$,so it is $Fe(II)O$.
$4$. $Fe_{2}O_{3}$: Oxidation state of $Fe$ is $+3$,so it is $Fe_{2}(III)O_{3}$.
$5$. $CuI$: Oxidation state of $Cu$ is $+1$,so it is $Cu(I)I$.
$6$. $CuO$: Oxidation state of $Cu$ is $+2$,so it is $Cu(II)O$.
$7$. $MnO$: Oxidation state of $Mn$ is $+2$,so it is $Mn(II)O$.
$8$. $MnO_{2}$: Oxidation state of $Mn$ is $+4$,so it is $Mn(IV)O_{2}$.