General Characteristics Questions in English

(B) The effective magnetic moment is calculated using the formula $\mu = \sqrt{n(n + 2)} \text{ B.M.}$,where $n$ is the number of unpaired electrons.
For $Sc^{3+}$,the electronic configuration is $[Ar] 3d^0$.
Since there are no unpaired electrons,$n = 0$.
Therefore,$\mu = \sqrt{0(0 + 2)} = 0 \text{ B.M.}$

(A) The oxidation state of $Cr$ in the given compounds is as follows:
$K_2CrO_4$: $Cr$ is in $+6$ oxidation state.
$CrO_2$: $Cr$ is in $+4$ oxidation state.
$CrCl_3$: $Cr$ is in $+3$ oxidation state.
$CrF_2$: $Cr$ is in $+2$ oxidation state.
As the positive oxidation state of a metal ion increases,the effective nuclear charge increases and the ionic radius decreases.
Therefore,$Cr$ in the $+6$ oxidation state $(K_2CrO_4)$ will have the smallest ionic radius.

(A) Isomorphic salts have the same crystal structure and similar chemical formulas. Since the copper salt is isomorphic with $ZnSO_4$,it is $CuSO_4$.
In $ZnSO_4$,the zinc ion is $Zn^{2+}$ $(3d^{10})$,which is diamagnetic.
However,the question asks for the nature of the copper salt $(CuSO_4)$.
In $CuSO_4$,the copper ion is $Cu^{2+}$.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
Since there is $1$ unpaired electron in the $3d$ orbital,the salt is paramagnetic in nature.

(D) $ZnO$ is white at room temperature.
On heating,it loses oxygen and forms non-stoichiometric $Zn_{1+x}O$,which appears yellow due to the presence of excess $Zn^{2+}$ ions in the interstitial sites.

(A) Invar is an alloy consisting of $64\% \ Fe$ and $36\% \ Ni$.
It is widely used in precision instruments like watches and meter scales because it possesses a very small coefficient of thermal expansion,meaning its dimensions remain nearly constant despite temperature changes.

(C) The correct answer is $(C)$.
In $CuCN$,the copper ion is $Cu^+$. The electronic configuration of $Cu^+$ is $[Ar] 3d^{10} 4s^0$.
Since all electrons in the $3d$ subshell are paired,there are no unpaired electrons available for $d-d$ transitions.
Therefore,$CuCN$ is expected to be colourless.

(C) Transition metal compounds exhibit colour in aqueous solution due to the presence of unpaired $d$-electrons,which allow for $d-d$ transitions.
In $Zn(NO_3)_2$,$Zn^{2+}$ has a $3d^{10}$ configuration (no unpaired electrons).
In $LiNO_3$,$Li^+$ is an alkali metal ion with no $d$-electrons.
In $Ba(NO_3)_2$,$Ba^{2+}$ is an alkaline earth metal ion with no $d$-electrons.
In $Co(NO_3)_2$,$Co^{2+}$ has a $3d^7$ configuration,which contains unpaired electrons,making the aqueous solution coloured.

(A) Transition metal compounds exhibit colour due to $d-d$ transitions,which require the presence of unpaired electrons.
Substances with unpaired electrons are paramagnetic.
Therefore,coloured compounds are generally paramagnetic.
Conversely,colourless compounds lack unpaired electrons,making them diamagnetic.
Thus,the statement that 'Colourless compounds of transition elements are paramagnetic' is incorrect.

(C) The colour of transition metal ions is due to the presence of unpaired electrons in the $d$-orbitals,which allow for $d-d$ transitions.
$(A)$ $ScCl_3$: $Sc^{3+}$ is $[Ar] 3d^0$,no unpaired electrons,colourless.
$(B)$ $TiO_2$: $Ti^{4+}$ is $[Ar] 3d^0$,no unpaired electrons,colourless.
$(C)$ $MnSO_4$: $Mn^{2+}$ is $[Ar] 3d^5$,has $5$ unpaired electrons,coloured.
$(D)$ $ZnSO_4$: $Zn^{2+}$ is $[Ar] 3d^{10}$,no unpaired electrons,colourless.
Therefore,$MnSO_4$ is coloured.

(B) An alloy is a homogeneous mixture of two or more metals or a metal and a non-metal.
$Fe$ and $Hg$ do not form an alloy (amalgam).
Iron and platinum are known to be resistant to forming amalgams with mercury $(Hg)$.

(B) Stainless steel does not rust because chromium forms an oxide layer and protects iron from rusting.

(B) The colour of transition metal ion salts is due to $d-d$ transitions of unpaired electrons in the $d-$orbital. Metal ion salts having the same number of unpaired electrons in their $d-$orbitals exhibit similar colours in an aqueous medium.
For $VOCl_2$,the metal ion is $V^{4+}$,which has the electronic configuration $[Ar] 3d^1$. It has $1$ unpaired electron.
For $CuCl_2$,the metal ion is $Cu^{2+}$,which has the electronic configuration $[Ar] 3d^9$. It also has $1$ unpaired electron.
Since both $V^{4+}$ and $Cu^{2+}$ have $1$ unpaired electron,they are expected to show similar colours.

(B) $Brass$ is an alloy of $Cu$ and $Zn$.
$German$ $silver$ is an alloy of $Cu$,$Zn$,and $Ni$.
Therefore,the common metal present in both is $Zn$.

(B) Brass contains $Cu$ $(60\%)$ and $Zn$ $(40\%)$ in its composition.

(B) The electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.
When $Fe$ loses two electrons,it forms $Fe^{2+}$ with a configuration of $[Ar] 3d^6$.
When $Fe$ loses three electrons,it forms $Fe^{3+}$ with a configuration of $[Ar] 3d^5$.
The $3d^5$ configuration is half-filled,which provides extra stability to the $Fe^{3+}$ ion.
Therefore,the $+3$ oxidation state is more stable than the $+2$ oxidation state.

(B) Permanent magnets are materials that retain their magnetic properties for a long time after the external magnetic field is removed.
Steel is a hard magnetic material with high retentivity and high coercivity,which makes it ideal for the preparation of permanent magnets.
In contrast,soft iron (wrought iron) is used for making electromagnets because it has high permeability and low retentivity,allowing it to be easily magnetized and demagnetized.

(C) Iron is a ferromagnetic material.
When heated above a specific temperature known as the Curie point $(T_C)$,the thermal energy becomes sufficient to overcome the exchange interactions that align the magnetic moments.
Consequently,iron loses its ferromagnetic property and becomes paramagnetic at temperatures above the Curie point.

(B) The correct option is $(B)$.
$Fe$ (Iron) is an essential element found in the human body,specifically as a central component of the haemoglobin molecule in red blood cells,which is responsible for oxygen transport.

(B) The correct answer is $(B)$.
$Mn$ (Manganese) is used to make alloy steel for armour plates,safes,and helmets because it provides high strength and resistance to wear.

(B) Copper is a transition metal and a very good conductor of electricity. Therefore,it is expected to have a high electrical conductivity,not a low one. The other properties listed,such as high thermal conductivity,ductility,and malleability,are characteristic properties of metals like copper.

(A) Transition metals often exhibit variable oxidation states,allowing them to form multiple compounds with the same non-metal.
Copper $(Cu)$ exhibits $+1$ and $+2$ oxidation states,forming $CuCl$ (cuprous chloride) and $CuCl_2$ (cupric chloride).
Other metals listed,such as $Al$,$Ag$,and $Na$,typically exhibit only one stable oxidation state in their chloride compounds ($AlCl_3$,$AgCl$,and $NaCl$ respectively).

Metal	Chlorides
$Cu$	$CuCl$ and $CuCl_2$
$Ag$	$AgCl$
$Na$	$NaCl$
$Al$	$AlCl_3$

(C) German silver is an alloy composed of $Cu$ (Copper),$Zn$ (Zinc),and $Ni$ (Nickel).
Therefore,besides $Zn$ and $Cu$,it contains $Ni$.

(C) $Cu$ is present in brass,bronze,and German silver.
All these substances are the alloys of copper.

(C) German silver consists of $Cu$,$Zn$,and $Ni$. This is correctly matched.
Alnico consists of $Fe$,$Al$,$Ni$,and $Co$. This is correctly matched.
Monel metal consists of $Ni$ and $Cu$ (often with small amounts of $Fe$ and $Mn$). It does not contain $Zn$ or $Sn$. Therefore,$Cu + Zn + Sn$ is incorrect; this composition actually represents Gun metal.
Duralumin consists of $Al$,$Cu$,$Mg$,and $Mn$. This is correctly matched.
Thus,the wrongly matched option is $C$.

(B) The electronic configuration of $Cu^{+}$ is $[Ar] 3d^{10}$. Since the $d$-orbital is completely filled,there are no unpaired electrons available for $d-d$ transition,making it colourless.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^{9}$. It has one unpaired electron in the $d$-orbital,which allows for $d-d$ transition,making it coloured.

(D) The atomic number of $Zn$ is $30$.
The electronic configuration of $Zn$ is $[Ar] 3d^{10} 4s^2$.
When $Zn$ forms $Zn^{2+}$ ion,it loses two electrons from the $4s$ orbital.
The electronic configuration of $Zn^{2+}$ is $[Ar] 3d^{10}$.
Since all $10$ electrons in the $3d$ subshell are paired,the number of unpaired electrons in $Zn^{2+}$ is $0$.

(B) The effective magnetic moment $(\mu_{eff})$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$,where $n$ is the number of unpaired electrons.
For $Cr^{2+}$: Electronic configuration is $[Ar] 3d^4$. Number of unpaired electrons $(n)$ = $4$.
For $Fe^{2+}$: Electronic configuration is $[Ar] 3d^6$. Number of unpaired electrons $(n)$ = $4$.
Since both $Cr^{2+}$ and $Fe^{2+}$ have $4$ unpaired electrons,they will have the same effective magnetic moment.
Therefore,the correct option is $B$.

(D) The number of unpaired electrons in each ion is calculated as follows:
$Mg^{2+} (Z=12): 1s^2, 2s^2 2p^6, 3s^0 \implies 0 \text{ unpaired electrons}$
$Ti^{3+} (Z=22): [Ar] 3d^1 \implies 1 \text{ unpaired electron}$
$V^{3+} (Z=23): [Ar] 3d^2 \implies 2 \text{ unpaired electrons}$
$Fe^{2+} (Z=26): [Ar] 3d^6 \implies 4 \text{ unpaired electrons}$
Thus,$Fe^{2+}$ has the maximum number of unpaired electrons.

(D) The correct answer is $(D)$.
$CoCl_3$ contains the $Co^{3+}$ ion.
The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.
It has $4$ unpaired electrons in the $3d$ subshell.
Due to the presence of unpaired electrons,$CoCl_3$ exhibits color,whereas $AgCl$,$HgCl_2$,and $ZnCl_2$ are white because they have a $d^{10}$ configuration with no unpaired electrons.

(C) The transition elements include $d$-block elements and $f$-block elements (inner transition elements).
Total number of transition and inner transition elements is approximately $61$ out of $118$ known elements.
Percentage = $\frac{\text{Number of transition elements}}{\text{Total number of elements}} \times 100$
Percentage = $\frac{61}{118} \times 100 \approx 51.69\%$.
Rounding to the nearest given option,the value is approximately $60\%$.

(B) To determine the paramagnetic nature,we calculate the number of unpaired electrons $(n)$ in each ion:
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$,$n=4$
$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^5$,$n=5$
$Cr^{3+}$ $(Z=24)$: $[Ar] 3d^3$,$n=3$
$Mn^{3+}$ $(Z=25)$: $[Ar] 3d^4$,$n=4$
Since $Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,it is the most paramagnetic.

(A) The basic character of metal oxides depends on the oxidation state and the metallic nature of the element.
For transition metal monoxides $(MO)$,the metal is in the $+2$ oxidation state.
As we move from left to right across the $3d$ series $(Ti$ $\rightarrow V$ $\rightarrow Cr$ $\rightarrow Fe)$,the electronegativity of the metal increases and its metallic character decreases.
Consequently,the ionic character of the $M-O$ bond decreases,leading to a decrease in the basic character of the oxides.
Therefore,the order of basic character is $TiO > VO > CrO > FeO$.

(B) To determine the degree of paramagnetism,we calculate the number of unpaired electrons $(n)$ in the metal ion of each compound:
$1$. In $MnSO_4 \cdot 4H_2O$,the metal ion is $Mn^{2+}$ $(3d^5)$,which has $n = 5$ unpaired electrons.
$2$. In $CuSO_4 \cdot 5H_2O$,the metal ion is $Cu^{2+}$ $(3d^9)$,which has $n = 1$ unpaired electron.
$3$. In $FeSO_4 \cdot 6H_2O$ and $FeSO_4 \cdot 5H_2O$,the metal ion is $Fe^{2+}$ $(3d^6)$,which has $n = 4$ unpaired electrons.
Since the degree of paramagnetism is directly proportional to the number of unpaired electrons,$CuSO_4 \cdot 5H_2O$ shows the lowest degree of paramagnetism.

(B) Substances that are diamagnetic are repelled by a magnetic field,causing them to appear to weigh less when weighed in such a field.
$ScCl_3$ dissociates as: $ScCl_3 \to Sc^{3+} + 3Cl^-$.
The electronic configuration of $Sc^{3+}$ is $[Ar] 3d^0 4s^0$.
Since there are no unpaired electrons,$Sc^{3+}$ is diamagnetic.
Therefore,$ScCl_3$ is repelled by the magnetic field and weighs less.

(A) The electronic configuration of the ion $M^{3+}$ is $[Ar]3d^1$.
To find the neutral atom $M$,we add $3$ electrons back to the ion.
$M = [Ar]3d^1 + 3e^- = [Ar]3d^2 4s^2$.
The atomic number of this element is $18 + 2 + 2 = 22$,which corresponds to Titanium $(Ti)$.
Therefore,the ion $M^{3+}$ is $Ti^{3+}$.

(B) In $Cu_2Cl_2$,the copper is in the $+1$ oxidation state $(Cu^+)$.
The electronic configuration of $Cu^+$ is $[Ar] 3d^{10}$.
Since the $3d$ subshell is completely filled,there are no unpaired electrons available for $d-d$ transitions.
Therefore,$Cu_2Cl_2$ is colourless.

(B) Diamond is an allotrope of $Carbon$.
Emerald is a variety of the mineral beryl $(Be_3Al_2Si_6O_{18})$ containing traces of chromium,which gives it its green color.
Therefore,Diamond is composed of $Carbon$ and Emerald is composed of $Alumina$ $(Al_2O_3)$ based silicates.

(A) In $CuF_2$,the copper ion is in the $+2$ oxidation state $(Cu^{2+})$. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. Due to the presence of an unpaired electron in the $d$-orbital,$d-d$ transitions are possible,which makes the compound colored.
In $CuI$,copper is in the $+1$ oxidation state $(Cu^+)$,which has a $3d^{10}$ configuration (no unpaired electrons).
$NaCl$ and $H_2O$ do not contain transition metal ions with partially filled $d$-orbitals,so they are colorless.

(D) To determine the number of unpaired electrons,we look at the electronic configuration of the metal ions:
$Fe^{+2}$ $(Z=26)$: $[Ar] 3d^6$ $\rightarrow$ $4$ unpaired electrons.
$Co^{+2}$ $(Z=27)$: $[Ar] 3d^7$ $\rightarrow$ $3$ unpaired electrons.
$Ni^{+2}$ $(Z=28)$: $[Ar] 3d^8$ $\rightarrow$ $2$ unpaired electrons.
$Mn^{+2}$ $(Z=25)$: $[Ar] 3d^5$ $\rightarrow$ $5$ unpaired electrons.
Thus,$Mn^{+2}$ has the maximum number of unpaired electrons $(5)$.

(A) The atomic number of $Fe$ is $26$. The electronic configuration of neutral $Fe$ is $[Ar] 3d^6 4s^2$.
When $Fe$ forms an $Fe^{2+}$ ion,it loses two electrons from the outermost $4s$ orbital.
Thus,the electronic configuration of $Fe^{2+}$ becomes $[Ar] 3d^6$.
The valence electrons are the electrons in the outermost shell or the partially filled $d$-subshell involved in bonding. For $Fe^{2+}$,the $3d$ electrons are considered valence electrons. There are $6$ electrons in the $3d$ subshell.

(D) The electronic configurations of the given ions are as follows:
$Fe^{2+} (Z=26) = [Ar] 3d^6$. Number of unpaired electrons = $4$.
$Co^{3+} (Z=27) = [Ar] 3d^6$. Number of unpaired electrons = $4$.
$Co^{2+} (Z=27) = [Ar] 3d^7$. Number of unpaired electrons = $3$.
$Fe^{3+} (Z=26) = [Ar] 3d^5$. Number of unpaired electrons = $5$.
Thus,$Fe^{3+}$ has the maximum number of unpaired electrons.

(D) The electronic configuration of $Mn^{2+}$ is $[Ar]3d^5$,which represents a stable half-filled $d$-orbital configuration.
Removing the third electron from $Mn^{2+}$ requires a significantly higher amount of energy because it involves disrupting this stable half-filled configuration.
Therefore,$Mn$ has the highest third ionization enthalpy among the given elements.

(B) The electronic configuration of the element with atomic number $Z = 106$ is $[Rn] 5f^{14} 6d^4 7s^2$.
Since the last electron enters the $d-$ orbital,the element belongs to the $d-$ block.

(C) The electronic configuration ends with the $3d$ subshell being filled.
Since the last electron enters the $d$-orbital,the element belongs to the $d$-block.

(D) The element $Lr$ (Lawrencium) is a synthetic (man-made) element,typically produced in particle accelerators. While $Fr$ (Francium) and $Rn$ (Radon) are naturally occurring radioactive elements,$Ra$ (Radium) is also a naturally occurring radioactive element. Therefore,$Lr$ is the correct choice among the options provided.

(C) The oxidation of a metal from the $+2$ to the $+3$ state is represented by the half-reaction: $M^{2+} \rightarrow M^{3+} + e^-$.
This process is favored when the reduction potential $E_{{M^{3+}}/{M^{2+}}}^{\circ}$ is low.
$A$ lower (or more negative) value of $E_{{M^{3+}}/{M^{2+}}}^{\circ}$ indicates that the $M^{2+}$ ion is more easily oxidized to the $M^{3+}$ ion.
Comparing the given values:
$Cr: -0.41 \ V$ (Note: The value for $Cr^{3+}/Cr^{2+}$ is actually negative,$-0.41 \ V$)
$Mn: +1.57 \ V$
$Fe: +0.77 \ V$
$Co: +1.97 \ V$
Since $Cr$ has the lowest (most negative) reduction potential,it is the most easily oxidized from the $+2$ to the $+3$ state.

(B) An ion is colored if it has unpaired electrons in its $d$-orbitals ($d-d$ transition).
$1$. $Sc^{3+}$ $([Ar] 3d^0)$: No unpaired electrons,colorless.
$2$. $Ti^{3+}$ $([Ar] 3d^1)$: One unpaired electron,colored.
$3$. $Ni^{2+}$ $([Ar] 3d^8)$: Two unpaired electrons,colored.
$4$. $Co^{3+}$ $([Ar] 3d^6)$: Four unpaired electrons,colored.
$5$. $Lu^{+}$ is not a standard transition metal ion in this context; assuming the question implies $Cu^{+}$ $([Ar] 3d^{10})$,it is colorless.
Comparing the options,$Ni^{2+}$ and $Ti^{3+}$ both have unpaired electrons and are colored in aqueous solution.

(D) compound is colored if it contains transition metal ions with partially filled $d$-orbitals,allowing for $d-d$ transitions.
$TiCl_3$ contains $Ti^{3+}$ $(3d^1)$,which is colored (purple).
$FeCl_3$ contains $Fe^{3+}$ $(3d^5)$,which is colored (yellow/brown).
$CoCl_2$ contains $Co^{2+}$ $(3d^7)$,which is colored (blue/pink).
Since all these compounds contain transition metal ions with unpaired electrons in $d$-orbitals,they are all colored.

(B) transition element is defined as an element which has an incompletely filled $d$-orbital in its ground state or in any of its oxidation states.
$Hg$ (Mercury) has the electronic configuration $[Xe] 4f^{14} 5d^{10} 6s^2$.
In its common oxidation state $(Hg^{2+})$,the configuration is $[Xe] 4f^{14} 5d^{10}$.
Since the $d$-orbital is completely filled in both the ground state and the common oxidation state,$Hg$ is not considered a transition element.

(D) The stability of transition metal ions in aqueous solution depends on their electronic configuration and hydration energy.
$Cr^{3+}$ has a $3d^3$ configuration,which corresponds to a half-filled $t_{2g}$ orbital in an octahedral field,providing extra stability.
$V^{3+}$ is $3d^2$,$Ti^{3+}$ is $3d^1$,and $Mn^{3+}$ is $3d^4$.
Among these,$Cr^{3+}$ is the most stable in aqueous solution due to the stable $t_{2g}^3$ configuration.