Let's analyze each option one by one:

Option A: The electronic configuration of Cr is [Ar] 3d⁵4s¹. (Atomic number of Cr = 24)

This statement is correct. Chromium (Cr) has an atomic number of 24, which means it has 24 electrons. The expected configuration, following the Aufbau principle, would be [Ar] 3d⁴4s². However, chromium is an exception to the expected order. To achieve a more stable and lower energy configuration, one electron from the 4s orbital moves to the 3d orbital, resulting in the configuration [Ar] 3d⁵4s¹. This configuration is more stable due to the half-filled 3d subshell, providing extra stability to the atom.

Option B: The magnetic quantum number may have negative value

This statement is correct. The magnetic quantum number, denoted as \( m_l \), describes the orientation of the orbital in space and can have integer values ranging from \(-l\) to \(+l\), where \(l\) is the azimuthal quantum number (also known as the orbital angular momentum quantum number). For example, if \(l = 2\) (d orbitals), \(m_l\) can have the values of -2, -1, 0, +1, +2, indicating that there are five different orientations for d orbitals in space.

Option C: In a silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of Ag = 47)

This statement is correct. Silver (Ag) has an atomic number of 47, indicating it has 47 electrons. Electrons are fermions, and they follow the Pauli exclusion principle, which states that no two electrons can have the same set of quantum numbers within an atom. This implies that electrons fill orbitals in pairs, with opposite spins. In the case of silver, with 47 electrons, this means there will be an unpaired electron. Thus, there will be 23 electrons with one spin direction and 24 electrons with the opposite spin, making the statement correct.

Option D: The oxidation state of Nitrogen in HN₃ is -3

This statement is incorrect. HN₃, also known as hydrazoic acid, contains nitrogen in a relatively unusual oxidation state. To determine the oxidation state of nitrogen in HN₃, let's assign oxidation numbers based on the usual oxidation states hydrogen (H) as \(+1\) and nitrogen (N), which will be determined. The sum of the oxidation states in a neutral compound is zero. In HN₃, with one H atom, hydrogen contributes \(+1\) to the total. If the nitrogen's oxidation state were \(-3\), the total would not balance to 0 with three nitrogen atoms. Instead, the structure and electronegativity of nitrogen suggest a different setup, where one nitrogen is more positive and the other two are more negative, but overall, nitrogen typically has positive oxidation states in HN₃. A common way to rationalize the states is to consider the central nitrogen in a covalent triple bond with a negative oxidation state, but overall, the nitrogen connected with hydrogen tends to have a positive oxidation state due to the high electronegativity of nitrogen compared to hydrogen. The correct total oxidation state of nitrogen in HN₃ depends on how the bonding is considered but is certainly not -3 overall for all nitrogen atoms.

Therefore, the correct statements are Option A and Option B, and Option C, while Option D is incorrect.