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Which of the sequences given below shows the correct increasing order of energy? (a) \(35,3 p, 4 s, 4 p, 3 d, 5 s, 5 p, 4 d\) (b) \(3 s, 3 p, 3 d, 4 s, 4 p, 4 d, 5 s, 5 p\) (c) \(3 s, 3 p, 4 s, 3 d, 4 p, 5 s, 4 d, 5 p\) (d) \(3 s, 3 p, 4 s, 4 p, 5 s, 3 d, 4 d, 5 p\)

Short Answer

Expert verified
The correct option showing the increasing order of energy is (c) 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p.

Step by step solution

01

Understanding the Aufbau Principle

To solve this exercise, use the Aufbau Principle which states that electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels. According to this principle, the correct order would be based on the increasing energy levels of the orbitals, taking into account the (n+l) rule where 'n' is the principal quantum number and 'l' is the azimuthal quantum number (orbital angular momentum quantum number).
02

Applying the (n+l) Rule

To determine the order, for each orbital, calculate the sum of 'n' and 'l'. Orbitals with a lower sum of 'n+l' are of lower energy. If two orbitals have the same 'n+l' value, the one with the lower 'n' value is of lower energy. Here are the calculated values: 3s (n=3, l=0, n+l=3), 3p (n=3, l=1, n+l=4), 4s (n=4, l=0, n+l=4), 3d (n=3, l=2, n+l=5), 4p (n=4, l=1, n+l=5), 5s (n=5, l=0, n+l=5), 4d (n=4, l=2, n+l=6), 5p (n=5, l=1, n+l=6).
03

Putting the Orbitals in Increasing Order

Once you have calculated the 'n+l' values for each orbital, put them in increasing order. If some orbitals have the same 'n+l' values, order them by their principal quantum number 'n'. The correct sequence is: 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p.
04

Identifying the Correct Option

Compare the sequence you've determined with the options given to find the one that matches. The sequence that conforms to the Aufbau Principle and is in the correct increasing order of energy is the sequence in option (c): 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Electron configuration refers to the distribution of electrons in the atomic orbitals of an atom. It follows a set of rules, where each electron is placed in the lowest energy orbital available, leading to the most stable arrangement. Think of it like a game where each electron needs to find the best seat in a theatre—the lower the energy 'seat', the better.

Each 'seat' or orbital can hold a certain number of electrons before it is considered filled: for s orbitals, the limit is 2; for p orbitals, it's 6; for d, it's 10; and for f, it's 14. The sequence of filling these orbitals is guided by the increasing energy levels, which can easily be tracked using a mnemonic or an energy level diagram. To exemplify, the electron configuration for carbon (with 6 electrons) in its ground state is 1s² 2s² 2p², indicating that the 1s orbital fills up first, followed by the 2s, and then moving on to the 2p. Understanding electron configuration is crucial as it informs the chemical properties and reactivity of the elements.
Atomic Orbitals
Atomic orbitals are regions within an atom where electrons are most likely to be found. They have different shapes and energy levels; the most common kinds are designated as s, p, d, and f. The s orbital is spherical, p orbitals are dumbbell-shaped, d orbitals are more complex, and f orbitals are even more so.

Each type of orbital has a set capacity for electrons—two for s orbitals, six for p orbitals, ten for d orbitals, and fourteen for f orbitals. Orbitals are not just boxes to be filled with electrons; they also have distinct energy levels which influence how they fill up according to the Aufbau Principle. Although each electron behaves like a wave and cannot have an exact location, atomic orbitals give us a practical way of understanding electron distribution and the structure of atoms.
Principal Quantum Number
The principal quantum number, denoted by the symbol 'n', essentially represents the 'shell' or energy level of an atomic orbital. It can have positive integer values (1, 2, 3,…). As 'n' increases, so does the orbital’s size and the energy level of electrons within them, which means electrons are further from the nucleus.

For example, the 1s orbital has a principal quantum number of 1 and is closer to the nucleus than the 2s orbital, which has a principal quantum number of 2. The larger the number, the higher the energy and the less tightly the electrons are held. This number is critical to determining the energy of an electron and the overall size of an atom.
Azimuthal Quantum Number
The azimuthal quantum number, often symbolized as 'l', indicates the subshell or type of orbital (s, p, d, f) and shapes the angular momentum of electrons. This number is integral to the energy level of an orbital and has values ranging from 0 to (n-1), for each value of n.

If 'n' is 1, then 'l' can only be 0, representing an s orbital. If 'n' is 2, 'l' can be 0 or 1, which correspond to s and p orbitals, respectively. The azimuthal quantum number helps us understand not just the shape but also the complexity of the electron configuration within an atom. For instance, when 'l' is 0, you have a spherical shape which is the simplest form. As 'l' increases, the shape and complexity of the orbitals grow, influencing how electrons are distributed and how atoms interact with each other.

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Most popular questions from this chapter

Effective nuclear charge \(\left(Z_{\text {eff }}\right)\) for a nucleus of an atom is defined as (a) shielding of the outermost shell electrons from the nucleus by the innermost shell electrons (b) the net positive charge experienced by electron from the nucleus (c) the attractive force experienced by the nucleus from electron (d) screening of positive charge on nucleus by innermost shell electrons.

The angular momentum of an electron in a given stationary state can be expressed as \(m_{e} v r=n \frac{h}{2 \pi}\). Based on this expression an electron can move only in those orbits for which its angular momentum is (a) equal to \(n\) (b) integral multiple of \(\frac{h}{2 \pi}\) (c) multiple of \(n\) (d) equal to \(\frac{h}{2 \pi}\) only.

What will be the energy of a photon which corresponds to the wavelength of \(0.50 \mathrm{~A}\) ? (a) \(3.98 \times 10^{-15} \mathrm{~J}\) (b) \(3 \times 10^{15} \mathrm{~J}\) (c) \(3.9 \times 10^{8} \mathrm{~J}\) (d) \(3 \times 10^{-34} \mathrm{~J}\)

Match the column I with column II and mark the appropriate choice. $$ \begin{array}{|l|l|l|l|} \hline {\begin{array}{c} \text { Column I } \\ \text { (Atom) } \end{array}} & &{\begin{array}{c} \text { Column II } \\ \text { (No. of unpaired electrons) } \end{array}} \\ \hline \text { (A) } & { }_{15} \mathrm{P} & \text { (i) } & \text { 6 unpaired electrons } \\ \hline \text { (B) } & { }_{24} \mathrm{Cr} & \text { (ii) } & \text { 2 unpaired electrons } \\ \hline \text { (C) } & { }_{26} \mathrm{Fe} & \text { (iii) } & \text { 3 unpaired electrons } \\ \hline \text { (D) } & { }_{14} \mathrm{Si} & \text { (iv) } & \text { 4 unpaired electrons } \\ \hline \end{array} $$ (a) (A) \(\rightarrow\) (ii), (B) \(\rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow\) (iii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (i), (B) \(\rightarrow\) (iii), (C) \(\rightarrow\) (ii), (D) \(\rightarrow\) (iv) (c) (A) \(\rightarrow\) (iii), (B) \(\rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow\) (iv), (D) \(\rightarrow\) (ii) (d) (A) \(\rightarrow\) (iv), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (i), (D) \(\rightarrow\) (iii)

The energy of the electron in a hydrogen atom has a negative sign for all possible orbits because (a) when the electron is attracted by the nucleus and is present in orbit \(n\), the energy is emitted and its energy is lowered. (b) when the electron is attracted by the nucleus and is present in orbit \(n\), the energy is absorbed and its energy is increased. (c) when the electron is repelled by the nucleus, the energy is released and its energy is lowered. (d) None of these.

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