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This is one of our many quizzes on Quantum mechanical model. Click on this to access more such competitive quizzes on Structure Of An Atom.

Quantum mechanical model Quiz 2

chemistry

 The electrons are filled in $s, p, d,$ and $f$ orbitals. So, how many electrons are there in the p-orbital of silicon

 Count all the electrons present in the p-orbital.

 Silicon is the element of the group $14$ and its atomic number is $14$ so, there will be $14$ electrons in the silicon atom. The configuration will be: $ \\ 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{2}} \\ $. In this configuration, there are $2$ p-orbitals that are: $2p$ and $3p$. In 2p there are $6$ electrons and in $3p$ there are $2$ electrons, so in total there are $8$ electrons in the p-orbital.

 $1{{s}^{1}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}4{{s}^{2}}$, this is an electronic configuration of an element in the ground state. What type of element is this?

  We can predict the type of element by checking the orbital in which the last electron enters.

The given configuration is $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}4{{s}^{2}}$, according to the increasing order of the energy of the orbitals, the configuration will be: $ \\ 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{1}} \\ $ The last electron is in the d-orbital of the $3d$ series, and all the elements of the $3d$ series are called the transition metals.

  Study the above diagram and identify the element: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7ecbc366-db49-4e28-acb7-cca6f39832f26328366610774869281.png"/>

 According to this diagram there are $22$ electrons and the element will come in the $4th$ group.

 The total number of electrons in the diagram is $22$, and the electronic configuration of $22$ electrons will be: $ \\ 1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{2}} \\ $ So, the element will be Titanium $(Ti)$.

 The specified orbital can be written in quantum numbers $n, l,$ and $m$. Here a set is given, find the correct orbital. $ \\ n = 5, l = 2, m = 0$

 The value of m is $0$ for ${{d}_{{{z}^{2}}}} , +1$ for ${{d}_{xz}} , -1$ for ${{d}_{yz}} , +2$ for ${{d}_{{{x}^{2}}-{{y}^{2}}}}$ , and -2 for ${{d}_{xy}}$.

The given information is $n = 5$, $l = 2$, $m = 0 \text{:} \\ $ The value of n is $5$, so the shell will be $5$. The value of l is $2$, so the subshell will be d. The value of m is $0$, so the orbit will be ${{d}_{{{z}^{2}}}}$. The orbital will be $5{{d}_{{{z}^{2}}}}$ 

There are two types of nodes in any orbital, i.e., radial / spherical nodes and planar/angular nodes. For $5p$, the radial and planar nodes are respectively.

 To calculate the radial nodes we can use the formula: $n – l – 1$. To calculate the planar nodes we can use the formula: $l$

 The given orbital is $5p$, in which the value of n will be $5$, and the value of $l$ will be $1. \\ $ To calculate the radial nodes we can use the formula: $n – l – 1. \\ $ The radial nodes will be = $5 – 1 – 1 = 3 \\ $ To calculate the planar nodes we can use the formula: $l \\ $ The planar nodes will be = $1$