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Chapter 2: Problem 110

Which of the following statement is false? (1) The orbit with more number of nodal planes will be of more energy. (2) The orbital with two angular nodes (nodal planes) is f-orbital. (3) The zero probability of finding the electron in \(\mathrm{p}_{x}-\) orbital is in \(y-z\) plane. (4) The orbital which do not has angular nodes is s-orbital.

Short Answer

Expert verified
Statement 2 is false.

Step by step solution

01

Understand Nodal Planes and Energy

Evaluate the statement 'The orbit with more number of nodal planes will be of more energy.' Correctly, the energy of an orbital increases with the number of angular nodes (nodal planes). Hence, this statement is true.
02

Examine Angular Nodes in f-orbitals

Review 'The orbital with two angular nodes (nodal planes) is f-orbital.' The f-orbitals actually have three angular nodes. Therefore, this statement is false.
03

Analyze Probability of Electron in \(\text{p}_x\)-orbital

Look at the statement 'The zero probability of finding the electron in \(\text{p}_x\)-orbital is in the \(y-z\) plane.' The \(\text{p}_x\)-orbital has a nodal plane at \(y = 0\) and \(z = 0\), which means this statement is true.
04

Check Angular Nodes in s-orbitals

Assess 'The orbital which does not have angular nodes is s-orbital.' s-orbitals indeed have no angular nodes, making this statement true.

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Key Concepts

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

Nodal Planes
Let's start by understanding nodal planes. Nodal planes are regions in an atom where the probability of finding an electron is zero. They are significant because they influence the energy of atomic orbitals. The greater the number of nodal planes, the higher the energy of the orbital.
Nodal planes are also known as angular nodes and differ from radial nodes, which are spherical regions where an electron cannot be found. Tilting a bit deeper:
  • Each type of orbital has a specific number of nodal planes.
  • For example, p-orbitals have one nodal plane, d-orbitals have two, and f-orbitals have three.
Understanding nodal planes helps us predict and explain the energy levels of electrons in different orbitals.
Angular Nodes
Angular nodes are the nodal planes we've discussed earlier. They are essential for determining the shape and energy of atomic orbitals. To dive in further:
  • Angular nodes depend on the orbital's angular momentum quantum number, denoted by the letter l.
  • For instance, a p-orbital (l = 1) has one angular node, a d-orbital (l = 2) has two, and an f-orbital (l = 3) has three.
  • These nodes are planes that cut through the nucleus and are where the electron density is zero.
Keep this in mind: the higher the number of angular nodes, the more complex the shape of the orbital and the higher its energy.
s-orbital
s-orbitals are the simplest type of atomic orbitals. Let's break down their characteristics:
  • They are spherical in shape and have no angular nodes. This means the probability of finding an electron is uniformly distributed around the nucleus.
  • Each energy level in an atom has one s-orbital, denoted as 1s, 2s, 3s, and so on.
  • Electrons in s-orbitals are found closer to the nucleus compared to p, d, or f orbitals.
S-orbitals play a crucial role in understanding electron configurations and are fundamental to the structure of atoms.

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Most popular questions from this chapter

The momentum of a photon is \(p\), the corresponding wavelength is (1) \(h / p\) (2) \(h p\) (3) \(p / h\) (4) \(h / \sqrt{p}\)

The electronic configuration of the element which is just above the element with atomic number 43 in the same group is (1) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{19} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{6}\) (2) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{5} 4 \mathrm{~s}^{2}\) (3) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{6} 4 \mathrm{~s}^{1}\) (4) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{10} 4 \mathrm{~s}^{2} 4 \mathrm{p}^{5}\)

Considering the three electronic transitions \(n=2 \rightarrow\) \(n=1, n=3 \rightarrow n=2\) and \(n=4 \rightarrow n=3\) for the hydrogen at which one of the following is true. (1) The photon emitted in the transition \(n=4\) to \(n=3\) would have the longest wavelength. (2) The photon emitted in the transition \(n=2\) to \(n=1\) would have the longest wavelength. (3) The transition from \(n=3\) to \(n=1\) is forbidden. (4) The electron does not experience any change in orbit radius for any of these transitions.

The ratio of the radius of the first Bohr orbit for the electron orbiting around the hydrogen nucleus that of the electron orbiting around the deuterium nucleus is approximately (1) \(1: 1\) (2) \(1: 2\) (3) \(2: 1\) (4) \(1: 4\)

Which electronic configuration does not follow Pauli's exclusion principle? (1) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4}\) (2) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{4} 4 \mathrm{~s}^{2}\) (3) \(1 s^{2} 2 p^{4}\) (4) \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{3}\)
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