General Characteristics Questions in English

(C) The atomic number of silver $(Ag)$ is $47$.
Following the Aufbau principle,the expected configuration would be $[Kr] 4d^{9} 5s^{2}$.
However,a completely filled $d$-subshell provides extra stability.
Therefore,one electron from the $5s$ orbital shifts to the $4d$ orbital to achieve the stable configuration $[Kr] 4d^{10} 5s^{1}$.

(C) The electronic configuration of $Fe$ $(Z = 26)$ is $1s^2, 2s^2 2p^6, 3s^2 3p^6, 4s^2 3d^6$.
From this configuration,it is evident that the orbitals present are $s, p,$ and $d$ orbitals.
Therefore,the correct option is $(C)$.

(B) The atomic number of copper is $29$.
According to the Aufbau principle,the expected configuration is $[Ar] 3d^9 4s^2$.
However,a fully-filled $3d$ subshell $(3d^{10})$ is more stable than a partially-filled one.
Therefore,one electron from the $4s$ orbital shifts to the $3d$ orbital to achieve the stable configuration $[Ar] 3d^{10} 4s^1$.

(A) The atomic number of the element is $Z = 106.$
Radon $(Rn)$ has an atomic number of $86.$
The electronic configuration is filled according to the Aufbau principle: $86 (Rn) + 14 (5f) + 4 (6d) + 2 (7s) = 106.$
Thus,the configuration is $[Rn] 5f^{14} 6d^4 7s^2.$

(B) The electronic configuration of $La$ $(Z=57)$ is $[Xe] 5d^1 6s^2$.
Following this,the $4f$ orbitals are filled.
Europium ($Eu$,$Z=63$) has a stable half-filled $4f$ subshell with the configuration $[Xe] 4f^7 6s^2$.
For Gadolinium ($Gd$,$Z=64$),the next electron enters the $5d$ orbital instead of the $4f$ orbital to maintain the stability of the half-filled $4f^7$ configuration.
Therefore,the electronic configuration of $Gd$ $(Z=64)$ is $[Xe] 4f^7 5d^1 6s^2$.

(A) 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.
Therefore,the electronic configuration of $Zn^{2+}$ is $[Ar] \, 3d^{10} 4s^0$.

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

(C) To determine the number of unpaired electrons,we look at the electronic configuration of the metal ions:
$Cu^{+}$ $(Z=29)$: $[Ar] 3d^{10}$,unpaired electrons = $0$.
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^{6}$,unpaired electrons = $4$.
$Fe^{3+}$ $(Z=26)$: $[Ar] 3d^{5}$,unpaired electrons = $5$.
$Co^{2+}$ $(Z=27)$: $[Ar] 3d^{7}$,unpaired electrons = $3$.
Therefore,the $Fe^{3+}$ ion has the highest number of unpaired electrons,which is $5$.

(A) The electronic configuration of the metal ion $X^{3+}$ is $[Ar]3d^4$.
This means the ion has $18$ (from $[Ar]$) $+ 4 = 22$ electrons.
Since the ion is formed by the loss of $3$ electrons from the neutral atom $X$,the number of electrons in the neutral atom is $22 + 3 = 25$.
Therefore,the atomic number $Z$ of the transition metal $X$ is $25$.

(D) $Mg^{2+}$ $(Z=12)$: $1s^2 2s^2 2p^6$ (No unpaired electrons).
$Ti^{3+}$ $(Z=22)$: $[Ar] 3d^1$ ($1$ unpaired electron).
$V^{3+}$ $(Z=23)$: $[Ar] 3d^2$ ($2$ unpaired electrons).
$Fe^{2+}$ $(Z=26)$: $[Ar] 3d^6$ ($4$ unpaired electrons).
Therefore,$Fe^{2+}$ has the maximum number of unpaired electrons.

(B) To determine the number of unpaired $d$-electrons,we look at the electronic configuration of each ion:
$Zn^{+}$ $(Z=30)$: $[Ar] \, 3d^{10} 4s^1$. Number of unpaired $d$-electrons = $0$.
$Fe^{2+}$ $(Z=26)$: $[Ar] \, 3d^6 4s^0$. The $3d^6$ configuration has $4$ unpaired electrons.
$N^{3+}$ $(Z=7)$: $1s^2 2s^2 2p^6$. This ion has no $d$-electrons.
$Cu^{+}$ $(Z=29)$: $[Ar] \, 3d^{10} 4s^0$. Number of unpaired $d$-electrons = $0$.
Comparing these,$Fe^{2+}$ has the most unpaired $d$-electrons $(4)$.

(B) The atomic number of cobalt $[Co]$ is $27$.
The electronic configuration of cobalt is $[Ar] \, 3d^7 4s^2$.
In the $3d$ subshell,there are $7$ electrons. According to Hund's rule,these electrons are filled as follows:
$3d^7$: The first $5$ electrons occupy the $5$ orbitals singly,and the remaining $2$ electrons pair up in the first two orbitals.
This leaves $3$ orbitals with a single electron each.
Therefore,there are $3$ unpaired electrons in cobalt metal.

(A) The atomic number of $Mn$ is $25$. The electronic configuration of $Mn$ is $[Ar] \, 3d^5 \, 4s^2$.
To form $Mn^{4+}$,we remove $4$ electrons ($2$ from $4s$ and $2$ from $3d$).
The electronic configuration of $Mn^{4+}$ is $[Ar] \, 3d^3$.
According to Hund's rule,the three electrons in the $3d$ subshell occupy separate orbitals,resulting in $3$ unpaired electrons.

(B) To determine the electronic configuration of the given species:
$1$. $Co^{3+}$ $(Z=27)$: The ground state configuration of $Co$ is $[Ar] 3d^{7} 4s^{2}$. Removing $3$ electrons gives $Co^{3+} = [Ar] 3d^{6} 4s^{0}$.
$2$. $Ni^{4+}$ $(Z=28)$: The ground state configuration of $Ni$ is $[Ar] 3d^{8} 4s^{2}$. Removing $4$ electrons gives $Ni^{4+} = [Ar] 3d^{6} 4s^{0}$.
Since both $Co^{3+}$ and $Ni^{4+}$ have the same electronic configuration $(3d^{6})$,option $(B)$ is correct.

(A) The element with atomic number $26$ is Iron $(Fe)$.
The electronic configuration of $Fe$ is $[Ar] 3d^6 4s^2$.
Due to the presence of unpaired electrons in the $3d$ orbitals,it exhibits magnetic properties.
Specifically,$Fe$ is a well-known ferromagnetic substance,which means it is strongly attracted by an external magnetic field and retains magnetism even after the field is removed.

(C) The electronic configuration of $Cu$ is $[Ar] \, 3d^{10} 4s^1$.
For $Cu^{+}$,the configuration is $[Ar] \, 3d^{10}$,which has no unpaired electrons (diamagnetic).
For $Cu^{2+}$,the configuration is $[Ar] \, 3d^9$,which has one unpaired electron.
Since $Cu^{2+}$ contains an unpaired electron,it is paramagnetic.
Therefore,the correct option is $(C)$.

(C) The colour of an electrolyte solution is determined by the presence of ions that can absorb specific wavelengths of light.
Often,the colour is associated with the presence of transition metal ions (cations) or specific complex anions.
Therefore,the colour depends on the nature of both the ions present in the solution.
For example,in $CuSO_4$ solution,the blue colour is due to the $Cu^{2+}$ ion,but in other salts,the anion may also contribute to the colour.

(C) The electronic configuration of $Cr$ $(Z = 24)$ is $[Ar] 3d^5 4s^1$.
Since $Cr$ has $6$ valence electrons ($5$ in $3d$ and $1$ in $4s$),it can lose all $6$ electrons to achieve a stable configuration.
Therefore,the maximum oxidation state of $Cr$ is $+6$,as seen in compounds like $K_2Cr_2O_7$ or $CrO_3$.

(A) The electronic configuration of iron $(Fe)$ is $[Ar] \, 3d^6 4s^2$.
When iron loses two electrons,it forms the ferrous ion $(Fe^{2+})$ with the configuration $[Ar] \, 3d^6$.
When iron loses three electrons,it forms the ferric ion $(Fe^{3+})$ with the configuration $[Ar] \, 3d^5$.
Therefore,the net charge on a ferrous ion is $+2$.

(D) . Many metals with catalytic properties because:
$(I)$. They provide a large surface area for reactions to occur.
$(II)$. They exhibit variable oxidation states.
$(III)$. They have vacant $d$-orbitals which allow them to form complexes and facilitate catalytic activity.

(C) The atomic number of the element is $29$.
The electronic configuration of this element $(Cu)$ is $[Ar] 3d^{10} 4s^{1}$.
Since the last electron enters the $d$-orbital,it belongs to the $d$-block.

(A) The general electronic configuration for $d$-block elements is $(n - 1)d^{1 - 10}ns^{0 - 2}$.
These elements are also known as transition elements because they exhibit properties intermediate between $s$-block and $p$-block elements.

(C) The electronic configuration is $[Kr] 4d^{10} 4f^{14} 5s^2 5p^6 5d^2 6s^2$.
In this configuration,the last electron enters the $5d$ subshell.
Elements in which the last electron enters the $d$-orbital of the penultimate shell are classified as $d$-block elements.
Therefore,the correct option is $C$.

(C) In a period of $d$-block transition metals,the atomic radii (and consequently atomic volumes) initially decrease due to the increase in effective nuclear charge. However,as more electrons are added to the $(n-1)d$ subshell,the screening effect increases,which balances the nuclear attraction. Consequently,the atomic radii remain nearly constant across the middle of the series before increasing slightly at the end due to electron-electron repulsions. Therefore,the overall trend is that they remain approximately the same.

(A) The atomic number of Radon $(Rn)$ is $86$.
For an element with atomic number $Z = 106$,the electronic configuration is filled as follows:
$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^{10}, 4p^6, 5s^2, 4d^{10}, 5p^6, 6s^2, 4f^{14}, 5d^{10}, 6p^6, 7s^2, 5f^{14}, 6d^4$.
However,according to the stability rules for $d$-block elements,the configuration is $[Rn] \, 5f^{14} \, 6d^4 \, 7s^2$.
Wait,checking the specific configuration for Seaborgium $(Z=106)$: The ground state configuration is $[Rn] \, 5f^{14} \, 6d^4 \, 7s^2$.

(C) The atomic number of the element is $21$.
The electronic configuration is $[Ar] \ 3d^1 4s^2$.
Since the last electron enters the $d$-orbital,it belongs to the $d$-block elements.

(B) Among all the elements,$Osmium$ $(Os)$ has the highest density,which is approximately $22.59 \, g/cm^3$.

(D) Transition metals form coloured salts due to the presence of partially filled $d$-orbitals,which allow for $d-d$ electronic transitions when they absorb light in the visible region.

(A) -block elements are known as transition elements because they exhibit properties intermediate between $s$-block and $p$-block elements.

(C) The electronic configuration of the element with atomic number $23$ (Vanadium) is $[Ar] \, 3d^3 \, 4s^2$.
Since the last electron enters the $d$-orbital,the element belongs to the $d$-block.

(B) The correct answer is $(B)$.
Among the given options,Tungsten $(W)$ is a metal that possesses the highest melting point of approximately $3422 \ ^\circ C$.
Note that Diamond is a non-metal (allotrope of carbon) and is not considered in the category of metals.

(A) The correct order for ionic radii of $M^{3+}$ ions is $Sc^{3+} > Cr^{3+} > Mn^{3+} > Fe^{3+}$.
Option $A$ is incorrect because the given order $Sc^{3+} > Cr^{3+} > Fe^{3+} > Mn^{3+}$ does not match the actual trend.
Option $B$ is correct as density generally increases across the $3d$ series.
Option $C$ is correct as ionic radii decrease with increasing atomic number for $M^{2+}$ ions,though there are minor variations.
Option $D$ is correct as basic nature depends on the oxidation state and electronegativity of the metal.

(C) The correct answer is $(C)$ $Fe, Co, Ni, Cu$.
In the $3d$ transition series,as we move from $Fe$ to $Cu$,the increase in nuclear charge is balanced by the shielding effect of the $d$-electrons.
As a result,the effective nuclear charge remains nearly constant,causing the atomic radii of these elements to be almost the same.

(D) The first ionisation energies of $Ti$,$V$,$Cr$,and $Mn$ are $656 \ kJ/mol$,$650 \ kJ/mol$,$652 \ kJ/mol$,and $717 \ kJ/mol$ respectively.
Ionisation energy generally increases from left to right in a period,although there are irregularities due to electronic configurations.
Among the given elements,$Mn$ $(3d^5 4s^2)$ has a stable half-filled $d$-subshell configuration,which contributes to its higher ionisation energy.
Therefore,$Mn$ has the maximum first ionisation potential.

(C) -block elements possess partially filled $(n-1)d$ orbitals. Because the energy difference between the $(n-1)d$ and $ns$ orbitals is small,electrons from both can participate in bonding,which allows these elements to exhibit variable valencies.

(D) The electronic configuration of cobalt ($Co$,$Z = 27$) is $[Ar] 3d^7 4s^2$.
In the $3d$ subshell,there are $7$ electrons. According to Hund's rule,these are filled as: $\uparrow\downarrow, \uparrow\downarrow, \uparrow, \uparrow, \uparrow$.
This results in $3$ unpaired electrons.
Since the atom contains unpaired electrons,it is paramagnetic.

(A) . Transition elements exhibit variable valency due to the presence of vacant $d$-orbitals and the small energy difference between $(n-1)d$ and $ns$ orbitals.

(B) The electronic configuration $3d^{10} 4s^1$ corresponds to Copper $(Cu)$,which is a transition metal.
The electronic configuration $4d^{10} 5s^1$ corresponds to Silver $(Ag)$,which is also a transition metal.
Both $Cu$ and $Ag$ belong to Group $11$ of the periodic table and are known as coinage metals.
Therefore,the correct option is $(B)$.

(C) $Fe$ (Iron) is a transition metal belonging to the $3d$ series. Transition metals are characterized by their ability to exhibit variable oxidation states due to the participation of both $(n-1)d$ and $ns$ electrons in bonding. $Fe$ commonly exhibits $+2$ and $+3$ oxidation states. In contrast,$Na$ $(+1)$,$Mg$ $(+2)$,and $Al$ $(+3)$ are representative elements that typically exhibit only one stable oxidation state.

(D) The electronic configurations of the given ions are as follows:
$Cu (Z=29) = [Ar] 3d^{10} 4s^1$. Therefore,$Cu^{+} = [Ar] 3d^{10}$,which has $18$ electrons in the outermost shell $(3s^2 3p^6 3d^{10})$.
$K (Z=19) = [Ar] 4s^1$. Therefore,$K^{+} = [Ar]$,which has $8$ electrons in the outermost shell.
$Cs (Z=55) = [Xe] 6s^1$. Therefore,$Cs^{+} = [Xe]$,which has $8$ electrons in the outermost shell.
$Th (Z=90) = [Rn] 6d^2 7s^2$. Therefore,$Th^{4+} = [Rn]$,which has $8$ electrons in the outermost shell.
Thus,$Cu^{+}$ is the ion that has $18$ electrons in its outermost shell.

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

(A) Compounds containing transition metal ions with unpaired $d$-electrons exhibit colour due to $d-d$ transitions.
In $CuF_2$,copper exists in the $Cu^{2+}$ oxidation state.
The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$.
Since it has one unpaired electron in the $3d$ orbital,it undergoes $d-d$ transitions and appears coloured.
In contrast,$CuI$ involves $Cu^+$ ($3d^{10}$,no unpaired electrons),while $NaCl$ and $MgCl_2$ contain ions with noble gas configurations ($d^0$ or $d^{10}$),making them colourless.

(C) The correct answer is $(C)$.
In $CrCl_3$,the chromium ion is $Cr^{3+}$.
The electronic configuration of $Cr^{3+}$ is $[Ar] 3d^3$.
Since it contains $3$ unpaired electrons in the $d$-orbital,it exhibits $d-d$ transitions,which results in the formation of coloured solutions.

(D) The atomic mass of the given elements are:
$Hg$ (Mercury) $\approx 200.59 \ u$
$Pb$ (Lead) $\approx 207.2 \ u$
$Ra$ (Radium) $\approx 226 \ u$
$U$ (Uranium) $\approx 238.03 \ u$
Comparing these values,$U$ (Uranium) has the highest atomic mass among the given options.
Therefore,$U$ is the heaviest metal.

(C) Radioactivity is a property of unstable nuclei.
$Cm$ (Curium),$No$ (Nobelium),and $Md$ (Mendelevium) are synthetic transuranic elements that are radioactive.
$Mo$ (Molybdenum) is a stable transition metal with atomic number $42$,and it is not radioactive.

(B) Transition metals are the most efficient catalysts because they possess partially filled $d-$orbitals,which allow them to exhibit variable oxidation states and provide a large surface area for adsorption.

(C) The correct answer is $(C)$.
Many of the $d-$block (transition) elements and their compounds act as efficient catalysts.
This catalytic property is primarily due to the availability of vacant $(n-1)d$ orbitals,the ability to exhibit variable oxidation states,and the formation of interstitial compounds.

(C) The atomic number of Chromium $(Cr)$ is $24$.
The electronic configuration of $Cr$ is $[Ar] 3d^5 4s^1$.
When $Cr$ forms $Cr^{+}$,one electron is removed from the $4s$ orbital.
Therefore,the electronic configuration of $Cr^{+}$ is $[Ar] 3d^5 4s^0$.
In the $3d^5$ subshell,there are $5$ unpaired electrons.
Thus,the number of unpaired electrons in $Cr^{+}$ is $5$.

(D) The electronic configuration of $Cr$ $(Z=24)$ is $[Ar] 3d^5 4s^1$.
Since $Cr$ has $6$ valence electrons ($5$ in $3d$ and $1$ in $4s$),it can lose all $6$ electrons to exhibit a maximum oxidation state of $+6$,as seen in compounds like $K_2Cr_2O_7$ or $CrO_3$.

(B) Transition elements show variable oxidation states because the energy difference between the $ns$ and $(n - 1)d$ orbitals is very small,allowing electrons from both to participate in bond formation.