
According to the CRC's, the melting point of cadmium bromide is ${868^ \circ }C$. Which of the following equations best describes what occurs when cadmium bromide is heated to ${868^ \circ }C$?
A) $CdB{r_{(s)}} \to CdB{r_{(l)}}$
B) $CdB{r_2}_{(s)} \to CdB{r_2}_{(l)}$
C) $CdB{r_{2(s)}} \to C{d_{\left( s \right)}} + B{r_{2(g)}}$
D) $CdB{r_{2(s)}} \to C{d^{2 + }}_{\left( {aq} \right)} + 2B{r^ - }_{(aq)}$
Answer
573.9k+ views
Hint: If a substances is heated up to its melting point the changes that occur in that substance are predominantly physical instead of chemical.
For example the molecules of the substance now will have more kinetic energy thus will be able to move freely in space with a higher kinetic energy.
Complete step by step solution:
Melting point of a solid substance is defined as the temperature at which its state gets converted into liquid from solid state
For example the melting point of ice is $273.15K$ At this temperature the ice which exist in a solid state gets converted to water which exist in liquid state
In the given question we have been provide with a halide of cadmium named as cadmium bromide
Now as we know that cadmium is a transition element of $4d$ series and has Atomic number $48$
It has a completely filled penultimate $d$ and outermost $s$ orbital $\left( {4{d^{10}}5{s^2}} \right)$ just like zinc $(3{d^{10}}4{s^2})$ metal hence shows $ + 2$ oxidation state predominantly
Therefore its compound with bromine which is an element of group $17$ called halogens will have formula $CdB{r_2}$
Now when Cadmium bromide will reach its Melting point the only change that would happen is physical
And that is
-Change in its physical state from solid into gaseous state.
And whatever heat is been provided thereafter will keep the temperature constant and will be utilized to free its molecules by reducing force of attraction between them by increasing their kinetic energy
Hence making change in cadmium bromide physical state
Hence the option “B”
$CdB{r_2}_{(s)} \to CdB{r_2}_{(l)}$
is the correct solution for the given question.
Note:
Even after raising the temperature of cadmium bromide up to its melting point the phase change will only occur after providing an extra amount of heat known as latent heat of fusion which is the heat required to convert a solid substance into its liquid state.
For example the molecules of the substance now will have more kinetic energy thus will be able to move freely in space with a higher kinetic energy.
Complete step by step solution:
Melting point of a solid substance is defined as the temperature at which its state gets converted into liquid from solid state
For example the melting point of ice is $273.15K$ At this temperature the ice which exist in a solid state gets converted to water which exist in liquid state
In the given question we have been provide with a halide of cadmium named as cadmium bromide
Now as we know that cadmium is a transition element of $4d$ series and has Atomic number $48$
It has a completely filled penultimate $d$ and outermost $s$ orbital $\left( {4{d^{10}}5{s^2}} \right)$ just like zinc $(3{d^{10}}4{s^2})$ metal hence shows $ + 2$ oxidation state predominantly
Therefore its compound with bromine which is an element of group $17$ called halogens will have formula $CdB{r_2}$
Now when Cadmium bromide will reach its Melting point the only change that would happen is physical
And that is
-Change in its physical state from solid into gaseous state.
And whatever heat is been provided thereafter will keep the temperature constant and will be utilized to free its molecules by reducing force of attraction between them by increasing their kinetic energy
Hence making change in cadmium bromide physical state
Hence the option “B”
$CdB{r_2}_{(s)} \to CdB{r_2}_{(l)}$
is the correct solution for the given question.
Note:
Even after raising the temperature of cadmium bromide up to its melting point the phase change will only occur after providing an extra amount of heat known as latent heat of fusion which is the heat required to convert a solid substance into its liquid state.
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