Physical properties Questions in English

(D) The densities of the given elements are as follows:
$Fe$ (Iron) $\approx 7.87 \, g/cm^3$
$Cu$ (Copper) $\approx 8.96 \, g/cm^3$
$B$ (Boron) $\approx 2.34 \, g/cm^3$
$Pb$ (Lead) $\approx 11.34 \, g/cm^3$
Comparing these values,$Pb$ has the highest density.
Therefore,the correct option is $(D)$.

(B) $Hg$ exhibits the lowest $M.P.$ and $B.P.$ among the given transition elements.
This is because $Hg$ has a completely filled $d$-orbital $([Xe] 4f^{14} 5d^{10} 6s^2)$,meaning there are no unpaired electrons available for metallic bonding.
Consequently,the interatomic forces are very weak,making $Hg$ a liquid at room temperature with a melting point of $234 \ K$.

(C) - Due to $d^5$ configuration,metallic bonds are weak.
$d^5$ orbital is half-filled,as a result,$3d$ electrons are more tightly held by the nucleus.
This reduces the delocalization of electrons,resulting in weaker metallic bonding.

(A) The boiling point of transition metals depends on the strength of metallic bonding,which is determined by the number of unpaired electrons in the $d$-orbitals.
$Zn$ has a completely filled $d$-orbital $(3d^{10} 4s^2)$,meaning it has no unpaired electrons available for metallic bonding.
Consequently,$Zn$ exhibits the weakest metallic bonding among the given options,resulting in the lowest boiling point.

(D) Density is defined as mass per unit volume $(d = \frac{m}{V})$.
Among the given elements,$Pb$ (Lead) has a very high atomic mass $(207.2 \ u)$.
Although $Pb$ has a larger atomic radius than $Fe$ or $Cu$,the significant increase in atomic mass outweighs the increase in volume,resulting in a higher density compared to the other options provided.

(C) Among the $d$-block elements,Tungsten $(W)$ has the highest melting point due to the strong metallic bonding resulting from the involvement of $d$-electrons in the formation of interatomic bonds.

(A) The correct answer is $A$. $Os$ and $Sc$.
Density is defined as mass per unit volume. Among the transition elements,$Osmium$ $(Os)$ has the highest density,which is approximately $22.60 \ g/cm^3$.
$Scandium$ $(Sc)$ has the lowest density among the $3d$ transition series elements,which is approximately $3.01 \ g/cm^3$.

(D) The enthalpy of atomization depends on the strength of metallic bonding.
In the $3d$ series,$Zn$ has a completely filled $d$-orbital $(3d^{10} 4s^2)$,which results in the absence of unpaired electrons for metallic bonding.
Consequently,$Zn$ exhibits the weakest metallic bonding,leading to the lowest enthalpy of atomization in the series.

(A) The melting point of transition metals depends on the strength of metallic bonding,which is influenced by the number of unpaired electrons available for bonding.
Among the given elements,Tungsten $(W)$ has the highest melting point $(3422 \ ^{\circ}C)$ because it has the maximum number of unpaired electrons in its $d$-orbitals,leading to the strongest metallic bonding.
Therefore,the correct option is $A$.

(A) The melting points of transition metals depend on the number of unpaired electrons and the strength of metallic bonding.
Generally,the melting point increases as the number of unpaired electrons increases,reaching a maximum in the middle of the series.
For the $3d$ series elements:
$Zn$ $([Ar] 3d^{10} 4s^2)$ has no unpaired electrons,resulting in weak metallic bonding and a low melting point $(419 \ ^\circ C)$.
$Cu$ $([Ar] 3d^{10} 4s^1)$ has one unpaired electron,resulting in a relatively low melting point $(1083 \ ^\circ C)$.
$Ni$ $([Ar] 3d^8 4s^2)$ has two unpaired electrons $(1455 \ ^\circ C)$.
$Fe$ $([Ar] 3d^6 4s^2)$ has four unpaired electrons,leading to stronger metallic bonding and a higher melting point $(1538 \ ^\circ C)$.
Therefore,the correct order of increasing melting point is $Zn < Cu < Ni < Fe$.

(A) In the $1^{st}$ transition series ($3d$ series),the maximum melting point is observed for $Cr$ $(Z=24)$ due to the stable $d^5$ configuration,which leads to strong metallic bonding. The minimum melting point is observed for $Zn$ $(Z=30)$ because it has a completely filled $d^{10}$ configuration,resulting in weak metallic bonding.
In the $2^{nd}$ transition series ($4d$ series),the maximum melting point is observed for $Mo$ $(Z=42)$ due to the stable $d^5$ configuration. The minimum melting point is observed for $Cd$ $(Z=48)$ due to the stable $d^{10}$ configuration.
Therefore,the maximum and minimum melting points for the $1^{st}$ series are $Cr$ and $Zn$,and for the $2^{nd}$ series are $Mo$ and $Cd$.

(N/A) Metallic properties: All transition elements exhibit typical metallic properties such as $(i)$ high tensile strength,$(ii)$ ductility,$(iii)$ malleability,and $(iv)$ high thermal and electrical conductivity.
$(b)$ Metallic structures: The lattice structures of transition elements are given in the table below:

Element	Structure
$Sc$	$hcp, bcc$
$Ti$	$hcp, bcc$
$V$	$bcc$
$Cr$	$bcc, ccp$
$Mn$	$X, bcc, ccp$
$Fe$	$bcc, hcp$
$Co$	$ccp, hcp$
$Ni$	$ccp$
$Cu$	$ccp$
$Zn$	$X, hcp$
$Y$	$hcp, bcc$
$Zr$	$hcp, bcc$
$Nb$	$bcc$
$Mo$	$bcc$
$Tc$	$hcp$
$Ru$	$hcp$
$Rh$	$ccp$
$Pd$	$ccp$
$Ag$	$ccp$
$Cd$	$X, hcp$
$La$	$hcp, ccp, bcc$
$Hf$	$hcp, bcc$
$Ta$	$bcc$
$W$	$bcc$
$Re$	$hcp$
$Os$	$hcp$
$Ir$	$ccp$
$Pt$	$ccp$
$Au$	$ccp$
$Hg$	$X$

Note: $bcc = \text{body-centered cubic}$,$hcp = \text{hexagonal close-packed}$,$ccp = \text{cubic close-packed}$,$X = \text{a specific metal structure}$. Except for $Zn, Cd, Hg, Mn$,transition elements exhibit one or more specific structures at room temperature.

(N/A) All lanthanoids are silvery-white soft metals and tarnish rapidly in air.
The hardness increases with an increase in atomic number. Samarium is steel-hard.
The melting points range between $1000 \ K$ and $1200 \ K$. However,samarium melts at $1623 \ K$.
They show luster and are good conductors of heat and electricity.
Density and other properties change smoothly except for $Eu$ and $Yb$,and occasionally for $Sm$ and $Tm$. Lanthanoids are paramagnetic,except for ions like $La^{3+}$,$Ce^{4+}$,$Yb^{2+}$,and $Lu^{3+}$.
The first ionization enthalpies of the lanthanoids are around $600 \ kJ \ mol^{-1}$ and the second is about $1200 \ kJ \ mol^{-1}$,which are comparable to those of calcium.
The third ionization enthalpies of elements lanthanum $(La)$,gadolinium $(Gd)$,and lutetium $(Lu)$ are found to be exceptionally low because of the stability of $f^{0}$,$f^{7}$,and $f^{14}$ shells,respectively,in these elements.

(A) Zinc $(Zn)$ and cadmium $(Cd)$ are relatively soft metals compared to other transition elements.
They form weaker metallic bonds due to the presence of completely filled $d$-orbitals ($d^{10}$ configuration),which do not participate in metallic bonding.
The melting point of $Zn$ is $419.5\,^{\circ}C$ and the melting point of $Cd$ is $321\,^{\circ}C$.
Both of these values are below $500\,^{\circ}C$.

(C) In the $3d$ transition series,density generally increases from left to right as atomic mass increases and atomic volume decreases.
However,$Zn$ has a lower density than $Cr$ because of its larger atomic size due to fully filled $d$-orbitals.
The density values (in $g/cm^3$) are: $Zn (7.13) < Cr (7.19) < Fe (7.87) < Co (8.90) < Cu (8.96)$.
Therefore,the correct order is $Zn < Cr < Fe < Co < Cu$.

(B) The melting points of transition elements are related to the strength of metallic bonding,which depends on the number of unpaired $d$-electrons available for bonding.
In the $3d$ series,$Cr$ $(3d^5 4s^1)$ has six unpaired electrons,leading to strong metallic bonding and a high melting point.
$Mn$ $(3d^5 4s^2)$ has a stable half-filled $d$-subshell,which makes the $d$-electrons less available for metallic bonding,resulting in a significantly lower melting point compared to $Cr$.
Therefore,the correct order is $Cr > Mn$.