Valency and oxidation state Questions in English

(N/A) $F$ exhibits only negative oxidation state of $-1$.
$(b)$ $Cs$ exhibits only positive oxidation state of $+1$.
$(c)$ $I$ exhibits both positive and negative oxidation states. It exhibits oxidation states of $-1, +1, +3, +5,$ and $+7$.
$(d)$ The oxidation state of $Ne$ is $0$. It exhibits neither negative nor positive oxidation states.

(A) Lithium $(Li)$ is in group-$1$ (valency $1$) and oxygen $(O)$ is in group-$16$ (valency $2$). The compound is $Li_{2}O$.
$(b)$ Magnesium $(Mg)$ is in group-$2$ (valency $2$) and nitrogen $(N)$ is in group-$15$ (valency $3$). The compound is $Mg_{3}N_{2}$.
$(c)$ Aluminium $(Al)$ is in group-$13$ (valency $3$) and iodine $(I)$ is in group-$17$ (valency $1$). The compound is $AlI_{3}$.
$(d)$ Silicon $(Si)$ is in group-$14$ (valency $4$) and oxygen $(O)$ is in group-$16$ (valency $2$). The compound is $SiO_{2}$.
$(e)$ Phosphorus $(P)$ is in group-$15$ (valency $5$) and fluorine $(F)$ is in group-$17$ (valency $1$). The compound is $PF_{5}$.
$(f)$ Element $71$ is Lutetium $(Lu)$,a lanthanoid with a stable oxidation state of $+3$. Fluorine $(F)$ has a valency of $1$. The compound is $LuF_{3}$.

Valency is defined as the combining capacity of an element. It represents the number of electrons an atom needs to lose,gain,or share to achieve a stable octet configuration in its outermost shell.
The valency of representative elements is usually equal to the number of electrons in the outermost orbitals or equal to $8$ minus the number of outermost electrons as shown below:

Group	$1$	$2$	$13$	$14$	$15$	$16$	$17$	$18$
Number of valence electrons	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Valency	$1$	$2$	$3$	$4$	$3$	$2$	$1$	$0$

(N/A) The valence or oxidation state is a characteristic property of elements,determined by their electronic configurations.
$1$. In $OF_2$,oxygen exhibits an oxidation state of $+2$,whereas in $Na_2O$,it exhibits $-2$.
$2$. The electronegativity order is $F > O > Na$.
$3$. In $OF_2$: Fluorine $(2s^2 2p^5)$ is the most electronegative element,so it is assigned an oxidation state of $-1$. Since there are two fluorine atoms,oxygen $(2s^2 2p^4)$ must share two electrons to complete its octet,resulting in an oxidation state of $+2$.
$4$. In $Na_2O$: Oxygen is more electronegative than sodium. Sodium $(3s^1)$ loses one electron to oxygen,acquiring an oxidation state of $+1$. Oxygen accepts two electrons (one from each sodium atom),resulting in an oxidation state of $-2$.

The valence of representative elements with respect to hydrogen and oxygen exhibits periodic trends across a period and down a group.
$1$. Valence with respect to hydrogen: It increases from $1$ to $4$ for groups $1$ to $14$ and then decreases from $4$ to $1$ for groups $14$ to $17$.
$2$. Valence with respect to oxygen: It generally increases from $1$ to $7$ across a period (equal to the group number for groups $1-13$ and $8$ minus the group number for groups $14-17$).

Group	$1$ to $13$	$14$	$15$	$16$	$17$
Formula of Hydride	$MH$ to $MH_3$	$MH_4$	$MH_3$	$MH_2$	$MH$
Formula of Oxide	$M_2O$ to $M_2O_3$	$MO_2$	$M_2O_5$	$MO_3$	$M_2O_7$

(D) $F$ (Fluorine) is the most electronegative element and exhibits only a $-1$ oxidation state in its compounds.
$(b)$ $Cs$ (Cesium) is an alkali metal and exhibits only a $+1$ oxidation state.
$(c)$ $I$ (Iodine) can exhibit negative $(-1)$ as well as positive $(+1, +3, +5, +7)$ oxidation states due to the presence of vacant $d$-orbitals.
$(d)$ $Ne$ (Neon) is a noble gas and has a zero oxidation state.
Therefore,all the given statements are correct.

The highest oxidation number of a representative element is equal to the group number for the first two groups,and the group number minus $10$ for the other groups.
Oxidation number generally increases from left to right across a period in the periodic table.

$Group$	$1$	$2$	$13$	$14$	$15$	$16$	$17$
$Element$	$Na$	$Mg$	$Al$	$Si$	$P$	$S$	$Cl$
$Compound$	$NaCl$	$MgSO_4$	$AlF_3$	$SiCl_4$	$P_4O_{10}$	$SF_6$	$HClO_4$
$Highest \ oxidation \ state$	$+1$	$+2$	$+3$	$+4$	$+5$	$+6$	$+7$

(B) The stock notation method is used to represent the oxidation state of a metal in a compound using Roman numerals in parentheses. This method is specifically applied to metals that exhibit more than one oxidation state,such as $Fe$ $(Fe^{2+}, Fe^{3+})$ or $Cu$ $(Cu^+, Cu^{2+})$.

(B) In $Cr_2O_3$,let the oxidation state of $Cr$ be $x$.
Since the oxidation state of $O$ is $-2$,we have:
$2(x) + 3(-2) = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the stock notation name is Chromium $(III)$ oxide.

(B) In $Fe_2(SO_4)_3$,the oxidation state of $Fe$ is calculated as follows:
Let the oxidation state of $Fe$ be $x$.
$2(x) + 3(-2) = 0$
$2x - 6 = 0$
$2x = 6$
$x = +3$
Therefore,the name according to the stock notation method is Iron $(III)$ sulphate.

(B) In $SnO_2$,the oxidation state of oxygen is $-2$. Let the oxidation state of $Sn$ be $x$.
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$.
Therefore,the stock notation name is Tin $(IV)$ oxide.

(C) Gallium $(Ga)$ belongs to Group $13$ of the periodic table.
Due to the inert pair effect,it exhibits both $+1$ and $+3$ oxidation states,with $+3$ being the most stable oxidation state.

(N/A) $(i)$ Aluminium $(Al)$ has a valency of $3$ (Group $13$) and Bromine $(Br)$ has a valency of $1$ (Group $17$). The formula is $AlBr_{3}$.
$(ii)$ Sodium $(Na)$ has a valency of $1$ (Group $1$) and Oxygen $(O)$ has a valency of $2$ (Group $16$). The formula is $Na_{2}O$.
$(iii)$ Silicon $(Si)$ has a valency of $4$ (Group $14$) and Oxygen $(O)$ has a valency of $2$ (Group $16$). The formula is $SiO_{2}$.
$(iv)$ Phosphorus $(P)$ has a valency of $5$ (Group $15$) and Chlorine $(Cl)$ has a valency of $1$ (Group $17$). The formula is $PCl_{5}$.

(A) Step $1$: Identify the elements.
$(i)$ The element with atomic number $19$ is Potassium $(K)$,which belongs to Group $1$. It has a valency of $+1$. Iodine $(I)$ belongs to Group $17$ (Halogens) and has a valency of $-1$. The compound formed is $KI$.
Step $2$: Identify the elements.
$(ii)$ The first element is Hydrogen $(H)$,which has a valency of $+1$. Oxygen $(O)$ belongs to Group $16$ and has a valency of $-2$. The compound formed is $H_2O$.

(N/A) The valency of carbon is $4$. When carbon forms three bonds and remains neutral,it is known as a carbon free radical. This happens because the carbon atom possesses one unpaired electron in its $p$-orbital,making it a neutral species despite having only three covalent bonds.

(C) In the compound $A_2 B$,the atom $B$ exists as $B^{-2}$ ion.
For an atom to achieve the lowest oxidation state of $-2$,it must gain $2$ electrons to complete its octet.
This implies that the atom $B$ has $6$ electrons in its valence shell in its neutral atomic state ($6 + 2 = 8$ electrons).
Therefore,the number of valence electrons in atom $B$ is $6$.

(D) The group valency is determined by the number of valence electrons or $8$ minus the number of valence electrons.
Elements in the second period $(N, O, F)$ cannot exhibit valencies matching their group number (e.g.,$N$ in group $15$ cannot show a valency of $5$) because they lack vacant $d$-orbitals in their valence shell.
$B, C, Al, Si, P, S$ can exhibit valencies corresponding to their group numbers.
Thus,the elements that cannot form compounds with valencies matching their group valencies are $N, O, F$.
The total count is $3$.

(B) The elements listed are $O, Cl, F, N, P, Sn, Tl, Na, Ti$.
$F$ (Fluorine) always shows $-1$ oxidation state.
$Na$ (Sodium) always shows $+1$ oxidation state.
Other elements show variable oxidation states:
$O$ $(-2, -1, +2)$,$Cl$ $(-1, +1, +3, +5, +7)$,$N$ ($-3$ to $+5$),$P$ ($-3$ to $+5$),$Sn$ $(+2, +4)$,$Tl$ $(+1, +3)$,$Ti$ $(+2, +3, +4)$.
Thus,only $2$ elements ($F$ and $Na$) show a single non-zero oxidation state.

(A) For the reaction $PbO_2 + Pb \longrightarrow 2 PbO,$ the value of $\Delta_{r} G^{\circ} < 0$ indicates that the reaction is spontaneous,meaning $PbO_2$ (where $Pb$ is in $+4$ state) readily converts to $PbO$ (where $Pb$ is in $+2$ state). Thus,$+2$ is the more stable and characteristic oxidation state for lead.
For the reaction $SnO_2 + Sn \longrightarrow 2 SnO,$ the value of $\Delta_{r} G^{\circ} > 0$ indicates that the reaction is non-spontaneous,meaning $SnO_2$ (where $Sn$ is in $+4$ state) is more stable than $SnO$ (where $Sn$ is in $+2$ state). Thus,$+4$ is the more stable and characteristic oxidation state for tin.
Therefore,the characteristic oxidation states are $+2$ for lead and $+4$ for tin.

(D) The chemical formula for manganese dioxide is $MnO_2$.
To find the oxidation state of $Mn$,let it be $x$.
The oxidation state of oxygen is $-2$.
So,$x + 2(-2) = 0$.
$x - 4 = 0$,which gives $x = +4$.
Therefore,the stock notation is $Mn(IV)O_2$.

(A) The maximum oxidation state of an element is generally related to the number of valence electrons.
For $Mn$ ($Z=25$,configuration: $[Ar] 3d^5 4s^2$),the maximum oxidation state is $+7$.
For $Cl$ ($Z=17$,configuration: $[Ne] 3s^2 3p^5$),the maximum oxidation state is $+7$ (e.g.,in $HClO_4$).
For $Cr$ $(Z=24)$,the maximum oxidation state is $+6$.
For $V$ $(Z=23)$,the maximum oxidation state is $+5$.
For $S$ $(Z=16)$,the maximum oxidation state is $+6$.
Therefore,the pair showing the highest oxidation state of $+7$ is $Mn$ and $Cl$.

(D) Group $1$ elements $(M)$ have an outermost shell electronic configuration of $ns^1$,making them monovalent with a valency of $+1$.
Group $16$ elements $(X)$ have an outermost shell electronic configuration of $ns^2 np^4$,requiring $2$ electrons to complete their octet,making them divalent with a valency of $-2$.
To form a neutral ionic compound,the total positive charge must balance the total negative charge.
Thus,two atoms of $M$ (each $+1$) are required to balance one atom of $X$ $(-2)$.
The resulting formula is $M_2 X$.

(A) The element with atomic number $117$ is $Ts$ (Tennessine).
The electronic configuration of $Ts$ is $[Rn] \ 5f^{14} \ 6d^{10} \ 7s^2 \ 7p^5$.
The valence shell is the $7^{th}$ shell,which contains $2 + 5 = 7$ electrons.
Therefore,the number of valence electrons is $7$.

(C) The electronic configuration of the element with $Z=13$ (Aluminum) is $[Ne] 3s^2 3p^1$.
The maximum covalency of an element is determined by the number of bonds it can form using its valence orbitals. For $Al$,it can expand its octet using vacant $3d$ orbitals to form complexes like $[AlF_6]^{3-}$,resulting in a maximum covalency of $6$.
Since the valence shell configuration is $3s^2 3p^1$,the element has $3$ valence electrons. Therefore,the highest oxidation state it can exhibit is $+3$.

(C) $K$ $(+1)$,$Sc$ $(+3)$,and $F$ $(-1)$ show only one non-zero oxidation state.
The oxidation states of $O$ are $-1, -2, +2$.
The oxidation states of $Cl$ range from $-1$ to $+7$.
The oxidation states of $N$ and $P$ range from $-3$ to $+5$.
The oxidation states of $Sn$ range from $+2$ to $+4$.
The oxidation states of $Ti$ are $+2, +3, +4$.
Thus,the elements showing more than one non-zero oxidation state are $O, Cl, N, P, Sn, Ti$.
Therefore,the total number of such elements is $6$.
The correct option is $C$.

(A) The basic strength of an oxoanion is inversely proportional to the oxidation state of the central atom $(Cl)$.
Higher oxidation state of the central atom leads to greater acidic nature and lower basic strength.
Calculating the oxidation state of $Cl$ in each species:
$ClO_2^{-}: x + 2(-2) = -1 \implies x = +3$
$ClO_3^{-}: x + 3(-2) = -1 \implies x = +5$
$ClO_4^{-}: x + 4(-2) = -1 \implies x = +7$
Since the order of oxidation states is $ClO_4^{-} (+7) > ClO_3^{-} (+5) > ClO_2^{-} (+3)$,the order of acidic nature is $ClO_4^{-} > ClO_3^{-} > ClO_2^{-}$.
Therefore,the order of basic strength is the reverse: $ClO_2^{-} > ClO_3^{-} > ClO_4^{-}$.
Thus,option $(A)$ is the correct answer.