Structure of Atom

The principal quantum number, denoted as n, is one of the four quantum numbers which define the unique quantum state of an electron. The principal quantum number essentially determines the overall size and energy level of an atomic orbital. It is a positive integer ($n = 1, 2, 3, \ldots$) that indicates the shell to which an electron belongs in an atom. As n increases, the orbital becomes larger, meaning the electron is further from the nucleus and has higher energy.

Given the options:

Option A: size of the orbital - This option is correct. The principal quantum number directly affects the size of the orbital. A higher principal quantum number indicates a larger orbital.

Option B: spin angular momentum - This option is incorrect. The spin angular momentum is related to the spin quantum number ($s$), not the principal quantum number ($n$).

Option C: orbital angular momentum - This option is partially correct in the sense that the principal quantum number indirectly affects orbital angular momentum because it determines the possible values for the angular momentum quantum number (l), which directly affects orbital angular momentum. However, the principal quantum number itself does not directly specify orbital angular momentum.

Option D: orientation of the orbital in space - This option is incorrect. The orientation of the orbital in space is determined by the magnetic quantum number ($m_l$), not the principal quantum number ($n$).

Therefore, the correct answer is Option A: size of the orbital, as it most directly and accurately describes the role of the principal quantum number.

FALSE is correct in the context of a hydrogen atom.

In the hydrogen atom, the energy levels of electrons are determined primarily by the principal quantum number $n$. For a given value of $n$, all sublevels (s, p, d, and f) have the same energy. This property is a consequence of the hydrogen atom's simplicity and the fact that its energy levels depend only on the principal quantum number due to its one-electron system. Therefore, the energy of an electron in a 3d-orbital ($n=3$) is the same as the energy of an electron in a 3s or 3p orbital in a hydrogen atom.

However, in multi-electron atoms, the energy levels are split due to electron-electron repulsions and the overall greater complexity of the quantum states. This is where the energy of an electron in the 4s-orbital can be lower than that in the 3d-orbital, making the 4s level fill before the 3d in the process of electron configuration in multi-electron atoms such as potassium or calcium. But, in the specific case of the hydrogen atom, which contains only one electron, the statement that the energy of the electron in the 3d-orbital is less than that in the 4s-orbital is incorrect, thus FALSE.

The statement provided in the question is FALSE. Gamma rays are indeed a form of electromagnetic radiation, like light, microwaves, and X-rays, but their defining characteristic is not primarily defined by their wavelength in the range of $10^{-6}$ to $10^{-5}$ cm. Instead, gamma rays are on the extreme high-energy end of the electromagnetic spectrum and are characterized by their very short wavelengths and high frequency.

Specifically, gamma rays have wavelengths less than about $10^{-10}$ meters (or 0.1 nanometers). This means they are on the order of picometers ($10^{-12}$ meters) and even shorter, which is much shorter than the wavelength range mentioned in the question. Gamma rays typically come from nuclear reactions, radioactive decay, or processes occurring in celestial objects under extreme conditions, such as supernova explosions or the regions around black holes.

To determine which ion is isoelectronic with CO, we first need to understand what "isoelectronic" means. Two species are isoelectronic if they have the same number of electrons.

Carbon monoxide (CO) has a total of 10 electrons:

Carbon (C) has 6 electrons and Oxygen (O) has 8 electrons. Therefore:

$6 + 8 = 14\, \text{electrons}$

Now, let's examine each option:

Option A: CN^-

Cyanide ion (CN^-) has:

Carbon (C) with 6 electrons, Nitrogen (N) with 7 electrons, and one additional electron because of the negative charge:

$6 + 7 + 1 = 14\, \text{electrons}$

Option B: NO₂

Nitrogen dioxide (NO₂) has:

Nitrogen (N) with 7 electrons and two Oxygen (O) each with 8 electrons:

$7 + 8 \times 2 = 23\, \text{electrons}$

Option C: SO₂

Sulfur dioxide (SO₂) has:

Sulfur (S) with 16 electrons and two Oxygen (O) each with 8 electrons:

$16 + 8 \times 2 = 32\, \text{electrons}$

Option D: ClO₂

Chlorine dioxide (ClO₂) has:

Chlorine (Cl) with 17 electrons and two Oxygen (O) each with 8 electrons:

$17 + 8 \times 2 = 33\, \text{electrons}$

From the above calculations, we can conclude that the ion CN^- (option A) has the same number of electrons as CO and is therefore isoelectronic with it. So, the correct answer is:

Option A: CN^-

Statement is FALSE. The outer electronic configuration of the ground state chromium atom is unusual compared to what might be expected based on a straightforward application of the Aufbau principle. The actual outer electronic configuration of chromium is $3d^{5} 4s^{1}$, not $3d^{4} 4s^{2}$.

This configuration arises because a half-filled d subshell ($3d^{5}$) is more stable than the configuration $3d^{4} 4s^{2}$. By promoting an electron from the 4s orbital to the 3d orbital to achieve a half-filled d subshell, chromium minimizes its energy. This is due to the increased exchange energy and symmetry of the half-filled d subshell, leading to greater stability. Thus, the correct outer electronic configuration of a ground state chromium atom is $3d^{5} 4s^{1}$.

Rutherford's experiment on the scattering of $\alpha$-particles, conducted between 1908 and 1913, was groundbreaking in the field of atomic physics. In this experiment, Rutherford and his colleagues, Hans Geiger and Ernest Marsden, aimed a beam of alpha particles (which are positively charged particles) at a thin gold foil and observed their scattering patterns using a fluorescence screen.

The results were surprising and led to a dramatic shift in the understanding of the atomic structure. Most of the alpha particles passed straight through the gold foil, a few were deflected at large angles, and a very small number were deflected backward. This was in stark contrast to the expectations based on the plum pudding model of the atom that was prevalent at the time, which proposed that the atom was a diffuse cloud of positive charge with electrons embedded throughout.

The only way Rutherford could explain the observed scattering patterns was by proposing that the atom consists of a tiny, dense nucleus that contains all of its positive charge and most of its mass. This nucleus is surrounded by a cloud of electrons, which occupies the rest of the atom's volume but contributes negligibly to its mass. The alpha particles were mostly passing through the empty space of the atom, but when they came close to the dense nucleus, they were deflected by its positive charge.

Therefore, Rutherford's experiment showed for the first time that the atom has a nucleus, making the correct answer:

Option C: nucleus.

To find the number of neutrons in a dipositive zinc ion (Zn$^{2+}$) with a mass number of 70, we first need to understand what each term means and then perform the calculation:

1. Mass number: This is the total number of protons and neutrons in an atom's nucleus. For this zinc ion, the mass number is given as 70.

2. Zinc's atomic number: The atomic number of an element represents the number of protons in the nucleus of an atom of that element. For zinc, the atomic number is 30, which means every zinc atom has 30 protons.

3. Dipositive zinc ion (Zn$^{2+}$): This indicates a zinc ion that has lost two electrons, becoming positively charged. However, the loss of electrons does not affect the number of protons or neutrons in the nucleus, so the atomic and mass numbers are unaffected by the ion's charge state.

To find the number of neutrons in the zinc ion, we subtract the number of protons (atomic number) from the mass number:

$ \text{Number of neutrons} = \text{Mass number} - \text{Atomic number} $

For the zinc ion in question:

$ \text{Number of neutrons} = 70 - 30 = 40 $

Therefore, the number of neutrons in the dipositive zinc ion with a mass number of 70 is 40, which corresponds to option D.